East China Normal University
Team’s Name
Chinese：道生一，一生二，二生三，三生万物
English：One,Two,Three,AK
Members
XuLiang Zhu - Mathematics
Mengyun Chen - Computer Science
Jiadong Xie - Software Engineering
文章目录
对拍
建立一个 test.sh 的文件
While true; do
./maker
./a
./b
if diff 1.out 2.out; then echo “AC”;else break;
fi
done
终端编译命令
chmod +x test.sh
./test.sh
输入输出优化
#define LL long long
using namespace std;
inline char nc(){
/*
static char buf[100000],*p1=buf,*p2=buf;
if (p1==p2) { p2=(p1=buf)+fread(buf,1,100000,stdin); if (p1==p2) return EOF; }
return *p1++;
*/return getchar();
}
inline void read(int &x){
char c=nc();int b=1;
for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1;
for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b;
}
inline void read(LL &x){
char c=nc();LL b=1;
for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1;
for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b;
}
inline void read(char &x){
for (x=nc();!(x=='('||x==')');x=nc());
}
inline int read(char *s)
{
char c=nc();int len=1;
for(;!(c=='('||c==')');c=nc()) if (c==EOF) return 0;
for(;(c=='('||c==')');s[len++]=c,c=nc());
s[len++]=' ';
return len-2;
}
int wt,ss[19];
inline void print(int x){
if (x<0) x=-x,putchar('-');
if (!x) putchar(48); else {
for (wt=0;x;ss[++wt]=x%10,x/=10);
for (;wt;putchar(ss[wt]+48),wt--);}
}
inline void print(LL x){
if (x<0) x=-x,putchar('-');
if (!x) putchar(48); else {for (wt=0;x;ss[++wt]=x%10,x/=10);for (;wt;putchar(ss[wt]+48),wt--);}
}
数据结构
hash
常用质数
1572869, 3145739, 6291469, 12582917, 25165843, 50331653
7881299347898369 原根 6
180143985094819841 原根 6
1945555039024054273 原根 5
4179340454199820289 原根 3
字符串哈希
不同长度的字符串哈希，需要加一维长度。
二维hash
#define seed 13131
#define seed1 1313
#define maxn 1005
typedef unsigned long long ull;
using namespace std;
ull p,Hash[maxn][maxn],Hash1[maxn][maxn];
char s[maxn][maxn],str[105][105];
int n,m,x,y;
ull getHash()
{
ull a,b=0;
for(int i=0;i<x;i++)
{
a=0;
for(int j=0;j<y;j++)
a=a*seed1+str[i][j];
b=b*seed+a;
}
return b;
}
int getAns()
{
ull base,a;
int ans=0;
base=1;
for(int i=0;i<y;i++)
base*=seed1;
for(int i=0;i<n;i++)
{
a=0;
for(int j=0;j<y;j++)
a=a*seed1+s[i][j];
Hash1[i][y-1]=a;
for(int j=y;j<m;j++)
Hash1[i][j]=Hash1[i][j-1]*seed1-s[i][j-y]*base+s[i][j];
}
base=1;
for(int i=0;i<x;i++)
base*=seed;
for(int i=y-1;i<m;i++)
{
a=0;
for(int j=0;j<x;j++)
a=a*seed+Hash1[j][i];
Hash[x-1][i]=a;
if(a==p) ans++;
for(int j=x;j<n;j++)
{
Hash[j][i]=Hash[j-1][i]*seed-Hash1[j-x][i]*base+Hash1[j][i];
if(Hash[j][i]==p) ans++;
}
}
return ans;
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&m);
for(int i=0;i<n;i++)
scanf("%s",s[i]);
scanf("%d%d",&x,&y);
for(int i=0;i<x;i++)
scanf("%s",str[i]);
p=getHash(); //hash str
printf("%d
",getAns()); //hash s
}
return 0;
}
并查集
int Find(int x)
{
return x==fa[x]?x:fa[x]=Find(fa[x]);
}
int main()
{
p=Find(id[x]),q=Find(id[y]);
fa[p]=q;
}
可持久化并查集
修改操作（镜像）
1 p q：合并p和q所在集合，如果已经在一个集合中，忽略此命令
2 p q：把p移动到q所在集合，如果已经在一个集合中，忽略此命令
3 p：输出p所在集合的元素个数和该集合所有元素之和
const int MAXN = 100010;
int n,m;
int father[MAXN];
int sum[MAXN],num[MAXN];
void init(){
for(int i=1;i<=n;i++){
father[i]=i+n;
father[i+n]=i+n;
sum[i+n]=i;
num[i+n]=1;
}
}
int find(int x){
return father[x]!=x?father[x]=find(father[x]):x;
}
int main(){
int mark,x,y;
while(scanf("%d%d",&n,&m)!=EOF){
init();
for(register int i=0;i<m;i++){
scanf("%d",&mark);
if(mark==1){
scanf("%d%d",&x,&y);
int fx=find(x),fy=find(y);
if(fx!=fy){
father[fx]=fy;
sum[fy]+=sum[fx];
num[fy]+=num[fx];
}
}else if(mark==2){
scanf("%d%d",&x,&y);
int fx=find(x),fy=find(y);
if(fx!=fy){
father[x]=fy;
sum[fy]+=x,sum[fx]-=x;
num[fy]++,num[fx]--;
}
}else{
scanf("%d",&x);
int fx=find(x);
printf("%d %d
",num[fx],sum[fx]);
}
}
}
return 0;
}
历史查询
1 a b 合并a,b所在集合
2 k 回到第k次操作之后的状态(查询算作操作)
3 a b 询问a,b是否属于同一集合，是则输出1否则输出0
#define N 2000005
int n,m,sz;
int root[N],ls[N],rs[N],v[N],deep[N];
void build(int &k,int l,int r){
if(!k)k=++sz;
if(l==r){v[k]=l;return;}
int mid=(l+r)>>1;
build(ls[k],l,mid);
build(rs[k],mid+1,r);
}
void modify(int l,int r,int x,int &y,int pos,int val){
y=++sz;
if(l==r){v[y]=val;deep[y]=deep[x];return;}
ls[y]=ls[x];rs[y]=rs[x];
int mid=(l+r)>>1;
if(pos<=mid)
modify(l,mid,ls[x],ls[y],pos,val);
else modify(mid+1,r,rs[x],rs[y],pos,val);
}
int query(int k,int l,int r,int pos){
if(l==r)return k;
int mid=(l+r)>>1;
if(pos<=mid)return query(ls[k],l,mid,pos);
else return query(rs[k],mid+1,r,pos);
}
void add(int k,int l,int r,int pos){
if(l==r){deep[k]++;return;}
int mid=(l+r)>>1;
if(pos<=mid)add(ls[k],l,mid,pos);
else add(rs[k],mid+1,r,pos);
}
int find(int k,int x){
int p=query(k,1,n,x);
return x==v[p]?p:find(k,v[p]);
}
int main(){
int f,k,a,b;
cin>>n>>m;
build(root[0],1,n);
for(register int i=1;i<=m;i++){
cin>>f;
if(f==1){
root[i]=root[i-1];
cin>>a>>b;
int p=find(root[i],a),q=find(root[i],b);
if(v[p]==v[q])continue;
if(deep[p]>deep[q])swap(p,q);
modify(1,n,root[i-1],root[i],v[p],v[q]);
if(deep[p]==deep[q])add(root[i],1,n,v[q]);
}else if(f==2){
cin>>k;root[i]=root[k];
}else if(f==3){
root[i]=root[i-1];
cin>>a>>b;
int p=find(root[i],a),q=find(root[i],b);
if(v[p]==v[q])cout<<1<<endl;
else cout<<0<<endl;
}
}
return 0;
}
带权并查集
#include<iostream>
#include<cstdio>
using namespace std;
const int N=1e5+5;
int fa[N],rnk[N];
void Init(int n)
{
for(int i=0;i<=n;i++)
{
fa[i]=i;
rnk[i]=0;
}
}
int Find(int x)
{
if(x==fa[x])return x;
int temp=fa[x];
fa[x]=Find(fa[x]);
rnk[x]=rnk[x]+rnk[temp];
return fa[x];
}
void Merge(int s,int x,int y)
{
int rx=Find(x);
int ry=Find(y);
if(rx==ry)return ;
fa[rx]=ry;
rnk[rx]=s+rnk[y]-rnk[x];
}
int main()
{
int n,m,q;
cin>>n>>m>>q; //人数，关系数，询问数
Init(n);
while(m--)
{
int a,b,c;
scanf("%d%d%d",&a,&b,&c); //a比b高c分
Merge(c,a,b);
}
while(q--)
{
int a,b;
scanf("%d%d",&a,&b); //想知道a号同学的分数比b号同学的分数高几分
int ra=Find(a);
int rb=Find(b);
if(ra!=rb)printf("-1
");
else printf("%d
",rnk[a]-rnk[b]);
}
return 0;
}
线段树
区间加乘，区间和
const int N=100005;
int n,m,a[4*N];
LL tr[4*N],add[4*N],mul[4*N],p;
void update(int x)
{
tr[x]=(tr[x<<1]+tr[x<<1|1])%p;
}
void build(int now,int l,int r)
{
mul[now]=1;add[now]=0;
if (l==r){tr[now]=a[l];mul[now]=1;return;}
int mid=l+r>>1;
build(now<<1,l,mid);
build(now<<1|1,mid+1,r);
update(now);
}
void pushdown(int now,int l,int r)
{
int mid=l+r>>1;
if (mul[now]!=1)
{
tr[now<<1]=tr[now<<1]*mul[now]%p;
tr[now<<1|1]=tr[now<<1|1]*mul[now]%p;
mul[now<<1]=mul[now<<1]*mul[now]%p;
mul[now<<1|1]=mul[now<<1|1]*mul[now]%p;
add[now<<1]=add[now<<1]*mul[now]%p;
add[now<<1|1]=add[now<<1|1]*mul[now]%p;
mul[now]=1;
}
if (add[now])
{
tr[now<<1]+=add[now]*(mid-l+1)%p;
if (tr[now]>=p) tr[now]-=p;
tr[now<<1|1]+=add[now]*(r-mid)%p;
if (tr[now]>=p) tr[now]-=p;
add[now<<1]+=add[now];
if (add[now]>=p) add[now]-=p;
add[now<<1|1]+=add[now];
if (add[now]>=p) add[now]-=p;
add[now]=0;
}
}
void qjmul(int now,int l,int r,int ll,int rr,LL val)
{
if (ll<=l&&r<=rr)
{
tr[now]=tr[now]*val%p;
mul[now]=mul[now]*val%p;
add[now]=add[now]*val%p;
return;
}
int mid=l+r>>1;
pushdown(now,l,r);
if (ll<=mid) qjmul(now<<1,l,mid,ll,rr,val);
if (rr>mid) qjmul(now<<1|1,mid+1,r,ll,rr,val);
update(now);
}
void qjadd(int now,int l,int r,int ll,int rr,LL val)
{
if (ll<=l&&r<=rr)
{
tr[now]=(LL)(tr[now]+(LL)(r-l+1)*val%p);
if (tr[now]>=p) tr[now]-=p;
add[now]+=val;
if (add[now]>=p) add[now]-=p;
return;
}
int mid=l+r>>1;
pushdown(now,l,r);
if (ll<=mid) qjadd(now<<1,l,mid,ll,rr,val);
if (rr>mid) qjadd(now<<1|1,mid+1,r,ll,rr,val);
update(now);
}
LL qjsum(int now,int l,int r,int ll,int rr)
{
if (ll<=l&&r<=rr) return tr[now];
int mid=l+r>>1; LL ans=0;
pushdown(now,l,r);
if (ll<=mid) ans+=qjsum(now<<1,l,mid,ll,rr); ans%=p;
if (rr>mid) ans+=qjsum(now<<1|1,mid+1,r,ll,rr); ans%=p;
return ans;
}
int main()
{
read(n),read(m),read(p);
for (int i=1;i<=n;i++)
read(a[i]);
build(1,1,n);
for (int i=1;i<=m;i++)
{
int opt,x,y,z;
read(opt);read(x);read(y);
if (opt==2) read(z),qjadd(1,1,n,x,y,z);
if (opt==1) read(z),qjmul(1,1,n,x,y,z);
if (opt==3) print(qjsum(1,1,n,x,y)%p),puts("");
}
return 0;
}
区间覆盖，询问
1 a:询问是不是有连续长度为 a 的空房间,有的话住进最左边
2 a b:将[a,a+b-1]的房间清空
struct ST
{
int l,r,lc,rc,lf,rf,f,flag;
}a[200010];
int n,m;
void build(int l,int r,int x)
{
a[x].l=l,a[x].r=r;a[x].flag=0;
a[x].lf=a[x].rf=a[x].f=r-l+1;
if (l==r) {a[x].lc=a[x].rc=0;return ;}
int mid=(l+r)>>1;
build(l,mid,x*2);
build(mid+1,r,x*2+1);
a[x].lc=x*2;a[x].rc=x*2+1;
}
void Pushdown(int x)
{
if (a[x].flag==1)
{
a[x].flag=0;
a[a[x].lc].f=a[a[x].lc].lf=a[a[x].lc].rf=0;
if (a[a[x].lc].r!=a[a[x].lc].l) a[a[x].lc].flag=1;
a[a[x].rc].f=a[a[x].rc].lf=a[a[x].rc].rf=0;
if (a[a[x].rc].r!=a[a[x].rc].l) a[a[x].rc].flag=1;
}
if (a[x].flag==-1)
{
a[x].flag=0;
a[a[x].lc].f=a[a[x].lc].lf=a[a[x].lc].rf=a[a[x].lc].r-a[a[x].lc].l+1;
if (a[a[x].lc].r!=a[a[x].lc].l) a[a[x].lc].flag=-1;
a[a[x].rc].f=a[a[x].rc].lf=a[a[x].rc].rf=a[a[x].rc].r-a[a[x].rc].l+1;
if (a[a[x].rc].r!=a[a[x].rc].l) a[a[x].rc].flag=-1;
}
}
int query(int k,int x)
{
if (a[x].lf>=k) return a[x].l;
Pushdown(x);
if (a[a[x].lc].f>=k) return query(k,a[x].lc);
if (a[a[x].lc].rf+a[a[x].rc].lf>=k) return a[a[x].lc].r-a[a[x].lc].rf+1;
return query(k,a[x].rc);
}
void change(int l,int r,int x,int z)
{
if (l<=a[x].l && r>=a[x].r)
{
if (a[x].l==a[x].r)
{
if (z==1) a[x].f=a[x].rf=a[x].lf=0;
else a[x].f=a[x].rf=a[x].lf=1;
}
else
{
if (z==1) a[x].flag=1,a[x].f=a[x].lf=a[x].rf=0;
else a[x].flag=-1,a[x].f=a[x].lf=a[x].rf=a[x].r-a[x].l+1;
}
return ;
}
Pushdown(x);
int mid=(a[x].l+a[x].r)>>1;
if (l<=mid) change(l,r,x*2,z);
if (r>mid) change(l,r,x*2+1,z);
a[x].f=max(a[a[x].lc].f,a[a[x].rc].f);
a[x].f=max(a[x].f,a[a[x].lc].rf+a[a[x].rc].lf);
if (a[a[x].lc].lf==a[a[x].lc].r-a[a[x].lc].l+1) a[x].lf=a[a[x].lc].lf+a[a[x].rc].lf;
else a[x].lf=a[a[x].lc].lf;
if (a[a[x].rc].rf==a[a[x].rc].r-a[a[x].rc].l+1) a[x].rf=a[a[x].rc].rf+a[a[x].lc].rf;
else a[x].rf=a[a[x].rc].rf;
}
int main()
{
read(n);read(m);
build(1,n,1);
int x,y;
while (m--)
{
read(x);
if (x==1)
{
read(y);
if (a[1].f<y) print(0),putchar('
');
else
{
int ans=query(y,1);
print(ans),putchar('
');
change(ans,ans+y-1,1,1);
}
}
else
{
read(x);read(y);
change(x,x+y-1,1,-1);
}
}
return 0;
}
李超树
#define maxn 50010
int T,n,m;
double ans;
char s[50];
struct ST
{
double k,b;
bool flag;
}tree[4*maxn];
double cross(double k1,double b1,double k2,double b2) //计算交点（需要注意斜率相等的情况）
{
return (b2-b1)/(1.0*(k1-k2));
}
void Insert(int x,int l,int r,double B,double K)
{
if (!tree[x].flag) {tree[x].k=K,tree[x].b=B,tree[x].flag=1;return;}
int mid=l+r>>1;
double f1=1.0*K*l+B,f2=1.0*tree[x].k*l+tree[x].b,f3=1.0*K*r+B,f4=1.0*tree[x].k*r+tree[x].b;
if (f1<=f2&&f3<=f4) return;
if (f1>=f2&&f3>=f4) {tree[x].k=K,tree[x].b=B;return;}
double xx=cross(K,B,tree[x].k,tree[x].b);
if (f1>=f2)
{
if (xx<=mid) Insert(x<<1,l,mid,B,K);
else Insert(x<<1|1,mid+1,r,tree[x].b,tree[x].k),tree[x].k=K,tree[x].b=B;
}
else
{
if (xx>mid) Insert(x<<1|1,mid+1,r,B,K);
else Insert(x<<1,l,mid,tree[x].b,tree[x].k),tree[x].k=K,tree[x].b=B;
}
}
void query(int x,int l,int r,int q)
{
if (tree[x].flag) ans=max(ans,1.0*tree[x].k*q+tree[x].b); //这里不能return，要走到底
if (l==r) return ;
int mid=l+r>>1;
if (q<=mid)query(x<<1,l,mid,q);else query(x<<1|1,mid+1,r,q);
}
int main()
{
n=50000;
read(T);
while (T--)
{
int x=read(s);double B,K;
if (s[1]=='P') scanf("%lf%lf",&B,&K),Insert(1,1,n,B-K,K); //一次函数第一天为B，公差为K
else read(x),ans=0,query(1,1,n,x),print((LL)floor(ans/100.0)),puts(""); //询问第x天函数max
}
return 0;
}
线段树的合并
求子树中有几个节点的权值小于根结点的权值
dfs，回溯的时候合并子树所构成的线段树
const int N=100010,M=N*20;
int n,a[N],ans[N],root[N],disc[N];
vector<int> v[N];
namespace Segment_Tree{
int tot;
struct node{int l,r,a,b,sum;}T[M];
void up(int x){T[x].sum=T[T[x].l].sum+T[T[x].r].sum;}
int build(int l,int r,int p){
int x=++tot;
T[x].a=l; T[x].b=r; T[x].sum=0;
if(l==r){T[x].sum=1;return x;}
int mid=(l+r)>>1;
if(p<=mid){T[x].l=build(l,mid,p);}
else{T[x].r=build(mid+1,r,p);}
return up(x),x;
}
int ask(int x,int l,int r){
if(!x)return 0;
if(l<=T[x].a&&T[x].b<=r)return T[x].sum;
int mid=(T[x].a+T[x].b)>>1,res=0;
if(l<=mid)res+=ask(T[x].l,l,r);
if(r>mid)res+=ask(T[x].r,l,r);
return res;
}
int merge(int x,int y){
if(!x||!y)return x^y;
T[x].l=merge(T[x].l,T[y].l);
T[x].r=merge(T[x].r,T[y].r);
return up(x),x;
}
void dfs(int x,int fx){
int res=0;
for(int i=0;i<v[x].size();i++){
int y=v[x][i];
if(y==fx)continue;
dfs(y,x);
res+=ask(root[y],a[x]+1,n);
root[x]=merge(root[x],root[y]);
}ans[x]=res;
}
}
int main(){
scanf("%d",&n);
for(int i=1;i<=n;i++)scanf("%d",&a[i]),disc[i]=a[i];
sort(disc+1,disc+n+1);
int m=unique(disc+1,disc+n+1)-disc-1;
for(int i=1;i<=n;i++)a[i]=lower_bound(disc+1,disc+m+1,a[i])-disc;
for(int i=2;i<=n;i++){
int x; scanf("%d",&x);
v[x].push_back(i); v[i].push_back(x);
}
for(int i=1;i<=n;i++)root[i]=Segment_Tree::build(1,n,a[i]);
Segment_Tree::dfs(1,1);
for(int i=1;i<=n;i++)printf("%d
",ans[i]);
return 0;
}
线段树合并与拆分
1:(0,l,r)表示将区间[l,r]的数字升序排序
2:(1,l,r)表示将区间[l,r]的数字降序排序
最后询问第q位置上的数字。
#include<iostream>
#include<queue>
using namespace std;
#define mid ((l+r)>>1)
#define ls t<<1,l,mid
#define rs t<<1|1,mid+1,r
#define ND 2000010
#define mxn 100010
int n,m,k,x,tt,val,i,last,op,L,R,now,nxt,LEFT,RIGHT;
struct data{
int l,r,tp,nxt;
}c[ND];
struct segment_tree{
struct data{
bool exi;
int rmax;
}a[mxn<<2];
void update(int t){
a[t]=(data){
a[t<<1].exi||a[t<<1|1].exi,
a[t<<1|1].rmax?a[t<<1|1].rmax:a[t<<1].rmax
};
}
void ins(int t,int l,int r){
if (l==r) a[t]=(data){k?1:0,k}; else
if (x<=mid) ins(ls),update(t); else
ins(rs),update(t);
}
int ask(int t,int l,int r){
if (l>x||(!a[t].exi)) return 0;
if (l==r) return a[t].rmax;
return (tt=ask(rs))?tt:(mid<=x?a[t<<1].rmax:ask(ls));
}
}T;
queue <int> q;
int newnode(){
int k=q.front();
q.pop();
return k;
}
struct node{
int l,r,sum;
}a[ND];
void clear(int t){
a[t]=a[0];
q.push(t);
}
void build(int t,int l,int r){
a[t].sum=1;
if (l==r) return;
if (val<=mid) build(a[t].l=newnode(),l,mid);
else build(a[t].r=newnode(),mid+1,r);
}
int merge(int t1,int t2){
if (t1==0||t2==0) return t1+t2;
a[t1].l=merge(a[t1].l,a[t2].l);
a[t1].r=merge(a[t1].r,a[t2].r);
a[t1].sum+=a[t2].sum;
clear(t2);
return t1;
}
void split(int t1,int t2,int k){
int tt=a[a[t1].l].sum;
if (k>tt) split(a[t1].r,a[t2].r=newnode(),k-tt); else swap(a[t2].r,a[t1].r);
if (k<tt) split(a[t1].l,a[t2].l=newnode(),k);
a[t2].sum=a[t1].sum-k;
a[t1].sum=k;
}
int ask(int t,int l,int r,int k){
if (l==r) return l;
int tt=a[a[t].l].sum;
if (k>tt) return ask(a[t].r,mid+1,r,k-tt);
else return ask(a[t].l,l,mid,k);
}
int main(){
for (i=1;i<=ND-10;i++)q.push(i);
cin>>n>>m;
for (x=1,last=ND-5;x<=n;last=k,x++){
cin>>val;
c[k=c[last].nxt=newnode()]=(data){x,x,0,0};
T.ins(1,1,n);
build(k,1,n);
}
while (m--){
cin>>op>>L>>R;
x=L;
LEFT=T.ask(1,1,n);
if (L==c[LEFT].l){
RIGHT=newnode();
swap(a[RIGHT],a[LEFT]);
swap(c[RIGHT],c[LEFT]);
c[now=LEFT]=(data){L,R,op,RIGHT};
} else{
c[now=newnode()]=(data){L,R,op,RIGHT=newnode()};
if (!c[LEFT].tp) split(LEFT,RIGHT,L-c[LEFT].l);
else split(LEFT,RIGHT,c[LEFT].r-L+1),swap(a[LEFT],a[RIGHT]);
c[RIGHT]=c[LEFT];
c[RIGHT].l=L;
c[LEFT].r=L-1;
c[LEFT].nxt=now;
}
for (nxt=RIGHT;nxt&&R>=c[nxt].r;last=nxt,nxt=c[nxt].nxt,c[last]=c[0]){
merge(now,nxt);
x=c[nxt].l,k=0;
T.ins(1,1,n);
}
c[now].nxt=nxt;
if (nxt!=0&&c[nxt].l<=R){
x=c[nxt].l,k=0;
T.ins(1,1,n);
if (c[nxt].tp) split(nxt,tt=newnode(),c[nxt].r-R);
else split(nxt,tt=newnode(),R-c[nxt].l+1),swap(a[nxt],a[tt]);
merge(now,tt);
x=c[nxt].l=R+1;
k=nxt;
T.ins(1,1,n);
}
x=L,k=now;
T.ins(1,1,n);
}
cin>>x;
now=T.ask(1,1,n);
if (c[now].tp) cout<<ask(now,1,n,c[now].r-x+1)<<"
";
else cout<<ask(now,1,n,x-c[now].l+1)<<"
";
return 0;
}
权值线段树中位数查询
离散化后一共有 m 个元素，支持以下操作:
1.ins c d e:插入一个离散化后是 c 的元素，对 cnt 的贡献为 d，对 sum 的贡献为 e。
2.ask:查询所有数字与中位数的差值的绝对值的和。
struct ST
{
int v[N];LL sum[N];
void ins(int c,int d,int e)
{
int a=1,b=m,x=1,mid;
while(1)
{
v[x]+=d,sum[x]+=e;
if(a==b)return;
x<<=1;
if(c<=(mid=(a+b)>>1))b=mid;else x|=1,a=mid+1;
}
}
LL ask()
{
if(v[1]<=1)return 0;
int a=1,b=m,mid,t,k=(v[1]+1)/2,x=1,cnt=0;LL ans=0;
while(a<b)
{
mid=(a+b)>>1,t=v[x<<=1];
if(k<=t)cnt+=v[x|1],ans+=sum[x|1],b=mid;
else cnt−=t,ans−=sum[x],k−=t,a=mid+1,x|=1;
}
return ans−sum[x]/v[x]*cnt;
}
};
二维线段树
1.将子矩阵x1,y1,x2,y2所有元素加上V
2.将子矩阵x1,y1,x2,y2所有元素设为V
3查询将子矩阵x1,y1,x2,y2的和，最大，小值
#define inf 0x3f3f3f3f
int ans,MX,MN;
#define lson L,mid,(rt<<1)
#define rson mid+1,R,(rt<<1|1)
struct point
{
int lc,rc,sum,mx,mn,add,set;
int mid(){return (lc+rc)>>1;}
};
point **tree;
void init(int r,int c)
{
tree=new point*[r+1];
for(int i=0;i<=r;i++)
{
tree[i]=new point[c*4];
}
}
void deal(int r,int c)
{
for(int i=0;i<=r;i++)
{
delete tree[i];
}
delete tree;
}
void pushup(int r,int rt)
{
tree[r][rt].sum=tree[r][rt<<1].sum+tree[r][rt<<1|1].sum;
tree[r][rt].mx=max(tree[r][rt<<1].mx,tree[r][rt<<1|1].mx);
tree[r][rt].mn=min(tree[r][rt<<1].mn,tree[r][rt<<1|1].mn);
}
void pushdown(int r,int rt)
{
int lg=tree[r][rt].rc-tree[r][rt].lc+1;
int rset,radd;
rset=tree[r][rt].set;
radd=tree[r][rt].add;
if(tree[r][rt].set)
{
tree[r][rt<<1].sum=rset*(lg-(lg>>1));
tree[r][rt<<1|1].sum=rset*(lg>>1);
tree[r][rt<<1].mx=rset;
tree[r][rt<<1].mn=rset;
tree[r][rt<<1|1].mx=rset;
tree[r][rt<<1|1].mn=rset;
tree[r][rt<<1].set=rset;
tree[r][rt<<1|1].set=rset;
tree[r][rt<<1].add=0;
tree[r][rt<<1|1].add=0;
tree[r][rt].set=0;
}
if(tree[r][rt].add)
{
tree[r][rt<<1].sum+=radd*(lg-(lg>>1));
tree[r][rt<<1|1].sum+=radd*(lg>>1);
tree[r][rt<<1].mx+=radd;
tree[r][rt<<1].mn+=radd;
tree[r][rt<<1|1].mx+=radd;
tree[r][rt<<1|1].mn+=radd;
tree[r][rt<<1].add+=radd;
tree[r][rt<<1|1].add+=radd;
tree[r][rt].add=0;
}
}
void build(int r,int L,int R,int rt)
{
tree[r][rt].add=0;
tree[r][rt].set=0;
tree[r][rt].lc=L;
tree[r][rt].rc=R;
if(L==R)
{
tree[r][rt].sum=0;
tree[r][rt].mx=0;
tree[r][rt].mn=0;
return ;
}
int mid=tree[r][rt].mid();
build(r,lson);
build(r,rson);
pushup(r,rt);
}
void update_set(int r,int L,int R,int rt,int v)
{
if(L==tree[r][rt].lc&&tree[r][rt].rc==R)
{
tree[r][rt].add=0;
tree[r][rt].set=v;
tree[r][rt].sum=(R-L+1)*v;
tree[r][rt].mx=v;
tree[r][rt].mn=v;
return ;
}
pushdown(r,rt);
int mid=tree[r][rt].mid();
if(R<=mid)update_set(r,L,R,rt<<1,v);
else if(L>mid)update_set(r,L,R,rt<<1|1,v);
else
{
update_set(r,lson,v);
update_set(r,rson,v);
}
pushup(r,rt);
}
void update_add(int r,int L,int R,int rt,int v)
{
if(L==tree[r][rt].lc&&tree[r][rt].rc==R)
{
tree[r][rt].add+=v;
tree[r][rt].sum+=(R-L+1)*v;
tree[r][rt].mn+=v;
tree[r][rt].mx+=v;
return ;
}
pushdown(r,rt);
int mid=tree[r][rt].mid();
if(R<=mid)update_add(r,L,R,rt<<1,v);
else if(L>mid)update_add(r,L,R,rt<<1|1,v);
else
{
update_add(r,lson,v);
update_add(r,rson,v);
}
pushup(r,rt);
}
void query(int r,int L,int R,int rt)
{
if(L==tree[r][rt].lc&&tree[r][rt].rc==R)
{
ans+=tree[r][rt].sum;
MX=max(MX,tree[r][rt].mx);
MN=min(MN,tree[r][rt].mn);
return ;
}
pushdown(r,rt);
int mid=tree[r][rt].mid();
if(R<=mid)query(r,L,R,rt<<1);
else if(L>mid)query(r,L,R,rt<<1|1);
else
{
query(r,lson);
query(r,rson);
}
}
int main()
{
int r,c,m,k,x1,x2,y1,y2,v;
while(~scanf("%d%d%d",&r,&c,&m))
{
init(r,c);
for(int i=1;i<=r;i++)
{
build(i,1,c,1);
}
while(m--)
{
scanf("%d",&k);
if(k==1)
{
scanf("%d%d%d%d%d",&x1,&y1,&x2,&y2,&v);
for(int i=x1;i<=x2;i++)
{
update_add(i,y1,y2,1,v);
}
}
else if(k==2)
{
scanf("%d%d%d%d%d",&x1,&y1,&x2,&y2,&v);
for(int i=x1;i<=x2;i++)
{
update_set(i,y1,y2,1,v);
}
}
else
{
ans=0;
MX=-inf;
MN=inf;
scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
for(int i=x1;i<=x2;i++)
{
query(i,y1,y2,1);
}
printf("%d %d %d
",ans,MN,MX);
}
}
deal(r,c);
}
return 0;
}
笛卡尔树
最大值为根的笛卡尔树
#define MAXN 1000010
const int inf=0x3f3f3f3f;
stack<int>st;
int T,n;
struct data
{
int val,sz;//val满足堆，sz子树大小
int l,r,par;//左孩子，右孩子，父亲
}t[MAXN];
void init()
{
for(int i=0;i<=n;i++)
t[i].l=0,t[i].r=0,t[i].par=0,t[i].sz=0;//初始化
t[0].val=inf;
while(!st.empty()) st.pop();
st.push(0);
}
void build()
{
for(int i=1;i<=n;i++)
{
while(!st.empty()&&t[st.top()].val<t[i].val)//从栈顶往栈底遍历
st.pop();
int par=st.top();
t[i].par=par;
t[i].l=t[par].r;
t[t[par].r].par=i;
t[par].r=i;
st.push(i);
}
}
int main()
{
read(T);
while(T--)
{
read(n);
init();
for(int i=1;i<=n;i++)
read(t[i].val);
build();
}
}
树状数组
树上二分（一个log）
void change(int i,int x)
{
for(;i<=n;i+=i&(-i)) c[i]+=x;
}
int query(int i)
{
int s=0;
for (;i;i-=i&(-i)) s+=c[i];
return s;
}
int qry(int x) //i是sum[i]<x的最后一位
{
int i=0,ii,k,t=0,tt;
for (k=tot>>1;k;k>>=1)
if ((tt=t+c[ii=(i|k)])<x){t=tt;i=ii;}
return i+1;
}
区间修改，区间询问
void add(LL p,LL x)
{
for(int i=p;i<=n;i+=i&-i)
sum1[i]+=x,sum2[i]+=x*p;
}
void range_add(LL l,LL r,LL x)
{
add(l,x),add(r+1,-x);
}
LL ask(LL p)
{
LL res=0;
for(int i=p;i;i-=i&-i) res+=(p+1)*sum1[i]-sum2[i];
return res;
}
LL range_ask(LL l, LL r)
{
return ask(r)-ask(l-1);
}
三维树状数组
单点增，区间和
const int maxn = 128 + 10;
int N;
int c[maxn][maxn][maxn];
int lowerbit(int x){return x & (-x);}
void add(int x, int y, int z, int v)
{
for(int i = x; i <= N; i += lowerbit(i))
for(int j = y; j <= N; j += lowerbit(j))
for(int k = z; k <= N; k += lowerbit(k))
c[i][j][k] += v;
}
long long sum(int x, int y, int z)
{
long long ret = 0;
for(int i = x; i > 0; i -= lowerbit(i))
for(int j = y; j > 0; j -= lowerbit(j))
for(int k = z; k > 0; k -= lowerbit(k))
ret += c[i][j][k];
return ret;
}
int main()
{
int op;
while(scanf("%d", &N) == 1)
{
while(scanf("%d", &op) && op != 3)
{
if(op == 1)
{
int x, y, z, k;
scanf("%d%d%d%d", &x, &y, &z, &k);
add(x+1, y+1, z+1, k);
}
else
{
int x1, y1, z1, x2, y2, z2;
scanf("%d%d%d%d%d%d", &x1, &y1, &z1, &x2, &y2, &z2);
x1++; y1++; z1++;
x2++; y2++; z2++;
printf("%I64d
", sum(x2, y2, z2)-sum(x2, y1-1, z2)-sum(x1-1, y2, z2)+sum(x1-1, y1-1, z2)
-(sum(x2, y2, z1-1)-sum(x2, y1-1, z1-1)-sum(x1-1, y2, z1-1)+sum(x1-1, y1-1, z1-1)));
}
}
}
return 0;
}
主席树
区间k大数
int T,n,m,s,k,h[100010],b[100010],d[100010];
struct ST
{
int lc,rc,sum;
}a[2000010];
int Hash(int x)
{
return lower_bound(b+1,b+1+k,x)-b;
}
void build(int l,int r,int x)
{
s++;x=s;
if (l==r) {a[x].lc=a[x].rc=0;return ;}
int mid=(l+r)>>1;
a[x].lc=s+1;build(l,mid,x);
a[x].rc=s+1;build(mid+1,r,x);
}
void change(int x,int y,int z,int xx,int l,int r) //新结点编号,,,对应结点编号,[l,r]
{
if (l==r) {a[x].sum=a[xx].sum+z;return ;}
int mid=(l+r)>>1;
if (y<=mid)
{
a[x].rc=a[xx].rc;
s++;a[x].lc=s;
change(s,y,z,a[xx].lc,l,mid);
}
else
{
a[x].lc=a[xx].lc;
s++;a[x].rc=s;
change(s,y,z,a[xx].rc,mid+1,r);
}
a[x].sum=a[a[x].lc].sum+a[a[x].rc].sum;
}
int query(int x,int y,int z,int l,int r)
{
if (l==r) return b[l];
if (a[a[y].lc].sum-a[a[x].lc].sum>=z) query(a[x].lc,a[y].lc,z,l,(l+r)>>1);
else query(a[x].rc,a[y].rc,z-(a[a[y].lc].sum-a[a[x].lc].sum),((l+r)>>1)+1,r);
}
int main()
{
read(T);
while (T--)
{
read(n);read(m);
memset(a,0,sizeof(a));
for (int i=1;i<=n;i++)
read(b[i]),d[i]=b[i];
sort(b+1,b+1+n);
k=unique(b+1,b+1+n)-b-1; //离散数据
s=0;
build(1,n,1);
h[0]=1;
for (int i=1;i<=n;i++)
{
s++;h[i]=s;
change(s,Hash(d[i]),1,h[i-1],1,n);
}
int x,y,z;
while(m--)
{
read(x);read(y);read(z);
print(query(h[x-1],h[y],z,1,n));putchar('
'); //区间[x,y]中的第z大数
}
}
return 0;
}
树上k大
询问(u,v,k)，回答u xor lastans和v这两个节点间第K小的点权
const int maxn=100010;
int n,m;
int a[maxn],b[maxn];
int pos[maxn],cnt;
struct node{int v,nxt;}E[maxn<<1];
int head[maxn],tot;
int fa[maxn],dep[maxn],top[maxn];
int size[maxn],son[maxn];
int lft[maxn<<5],rht[maxn<<5];
int rt[maxn<<5],sum[maxn<<5],sz;
int ans;
void add(int u,int v)
{
E[++tot].nxt=head[u];
E[tot].v=v;
head[u]=tot;
}
void dfs1(int u,int pa)
{
size[u]=1;
for(int i=head[u];i;i=E[i].nxt)
{
int v=E[i].v;
if(v==pa) continue;
dep[v]=dep[u]+1; fa[v]=u;
dfs1(v,u);
size[u]+=size[v];
if(size[v]>size[son[u]]) son[u]=v;
}
}
void dfs2(int u,int tp)
{
top[u]=tp;
if(son[u]) dfs2(son[u],tp);
for(int i=head[u];i;i=E[i].nxt)
{
int v=E[i].v;
if(v==fa[u]||v==son[u]) continue;
dfs2(v,v);
}
}
int update(int pre,int ll,int rr,int x)
{
int rt=++sz;
lft[rt]=lft[pre]; rht[rt]=rht[pre]; sum[rt]=sum[pre]+1;
if(ll<rr)
{
int mid=ll+rr>>1;
if(x<=mid) lft[rt]=update(lft[pre],ll,mid,x);
else rht[rt]=update(rht[pre],mid+1,rr,x);
}
return rt;
}
void dfs(int u)
{
int x=lower_bound(pos+1,pos+1+cnt,a[u])-pos;
rt[u]=update(rt[fa[u]],1,cnt,x);//以fa[u]为上一棵主席树建树
for(int i=head[u];i;i=E[i].nxt)
if(E[i].v!=fa[u]) dfs(E[i].v);
}
int LCA(int u,int v)
{
while(top[u]!=top[v])
{
if(dep[top[u]]>dep[top[v]]) u=fa[top[u]];
else v=fa[top[v]];
}
if(dep[u]<dep[v]) return u;
else return v;
}
int query(int u,int v,int lca,int gra,int ll,int rr,int k)
{
if(ll==rr) return ll;
int x=sum[lft[u]]+sum[lft[v]]-sum[lft[lca]]-sum[lft[gra]];//差分得路径
int mid=ll+rr>>1;
if(x>=k) return query(lft[u],lft[v],lft[lca],lft[gra],ll,mid,k);
else return query(rht[u],rht[v],rht[lca],rht[gra],mid+1,rr,k-x);
}
int main()
{
scanf("%d%d",&n,&m);
for(int i=1;i<=n;++i)
scanf("%d",&a[i]),b[i]=a[i];
sort(b+1,b+1+n);
for(int i=1;i<=n;++i)
if(i==1||b[i]!=b[i-1])
pos[++cnt]=b[i];
for(int i=1;i<n;++i)
{
int u,v;scanf("%d%d",&u,&v);
add(u,v); add(v,u);
}
dfs1(1,0); dfs2(1,1);//树剖
dfs(1);//深搜建立主席树
while(m--)
{
int u,v,k;scanf("%d%d%d",&u,&v,&k);
u^=ans;
int lca=LCA(u,v);
ans=pos[query(rt[u],rt[v],rt[lca],rt[fa[lca]],1,cnt,k)];//差分得到u到v的路径
printf("%d
",ans);
}
return 0;
}
带修改的k大数
1 x y：把a[x]修改为y。
2 x y k：查询[x,y]内第k小值。
const int N = 200010, M = 524289;
int n, m, cnt, i, a[N], op[N][4], b[N];
int lower(int x)
{
int l = 1, r = cnt, t, mid;
while (l <= r)
if (b[mid = (l + r) >> 1] <= x)
l = (t = mid) + 1;
else
r = mid - 1;
return t;
}
struct node
{
int val, cnt, sum, p;
node *l, *r;
node()
{
val = sum = cnt = p = 0;
l = r = NULL;
}
inline void up() { sum = l->sum + r->sum + cnt; }
} *blank = new (node), *T[M], pool[2000000], *cur;
void Rotatel(node *&x)
{
node *y = x->r;
x->r = y->l;
x->up();
y->l = x;
y->up();
x = y;
}
void Rotater(node *&x)
{
node *y = x->l;
x->l = y->r;
x->up();
y->r = x;
y->up();
x = y;
}
void Ins(node *&x, int y, int p)
{
if (x == blank)
{
x = cur++;
x-> val = y;
x-> l = x-> r = blank;
x-> sum = x-> cnt = 1;
x-> p = std::rand();
return;
}
x-> sum += p;
if (y == x-> val)
{
x-> cnt += p;
return;
}
if (y<x-> val)
{
Ins(x-> l, y, p);
if (x-> l-> p > x-> p)
Rotater(x);
}
else
{
Ins(x-> r, y, p);
if (x-> r-> p > x-> p)
Rotatel(x);
}
}
int Ask(node *x, int y)
{ //ask how many <= y
int t = 0;
while (x != blank)
if (y<x-> val)
x = x-> l;
else
t += x-> l-> sum + x-> cnt, x = x-> r;
return t;
}
void add(int v, int i, int p)
{
int a = 1, b = cnt, mid, f = 1, x = 1;
while (a < b)
{
if (f)
Ins(T[x], i, p);
mid = (a + b) >> 1;
x <<= 1;
if (f = v <= mid)
b = mid;
else
a = mid + 1, x |= 1;
}
Ins(T[x], i, p);
}
int kth(int l, int r, int k)
{
int x = 1, a = 1, b = cnt, mid;
while (a < b)
{
mid = (a + b) >> 1;
x <<= 1;
int t = Ask(T[x], r)-Ask(T[x], l-1);
if (k <= t)
b = mid;
else
k-= t, a = mid + 1, x |= 1;
}
return a;
}
void build(int x, int a, int b)
{
T[x] = blank;
if (a == b)
return;
int mid = (a + b) >> 1;
build(x << 1, a, mid), build(x << 1 | 1, mid + 1, b);
}
int main()
{
int T;read(T);
blank-> l = blank-> r = blank;
while (T--)
{
read(n);read(m);
cur = pool;
for (i = 1; i <= n; i++)
read(a[i]), b[i] = a[i];
cnt = n;
for (i = 1; i <= m; i++)
{
read(op[i][0]), read(op[i][1]), read(op[i][2]);
if (op[i][0] == 1)
b[++cnt] = op[i][2];
else
read(op[i][3]);
}
sort(b + 1, b + cnt + 1);
for (i = 1; i <= n; i++)
a[i] = lower(a[i]);
for (i = 1; i <= m; i++)
if (op[i][0] == 1)
op[i][2] = lower(op[i][2]);
build(1, 1, cnt);
for (i = 1; i <= n; i++)
add(a[i], i, 1);
for (i = 1; i <= m; i++)
{
if (op[i][0] == 1)
add(a[op[i][1]], op[i][1],-1), add(a[op[i][1]] = op[i][2], op[i][1], 1);
else
printf("%d
", b[kth(op[i][1], op[i][2], op[i][3])]);
}
}
}
区间不同数的个数
const int maxn = 30000 + 10;
int n,q,cnt = 0,la[1000000 + 10],a[maxn],root[maxn];
struct Node
{
int l,r,sum;
}p[maxn*40];
int build(int l,int r)
{
int nc = ++cnt;
p[nc].sum = 0;
p[nc].l = p[nc].r = 0;
if (l == r) return nc;
int m = l + r >> 1;
p[nc].l = build(l,m);
p[nc].r = build(m+1,r);
return nc;
}
int update(int pos,int c,int v,int l,int r)
{
int nc = ++cnt;
p[nc] = p[c];
p[nc].sum += v;
if (l == r) return nc;
int m = l+r>>1;
if (m >= pos) p[nc].l = update(pos,p[c].l,v,l,m);
else p[nc].r = update(pos,p[c].r,v,m+1,r);
return nc;
}
int query(int pos,int c,int l,int r)
{
if (l == r) return p[c].sum;
int m = l + r >> 1;
if (m >= pos) return p[p[c].r ].sum + query(pos,p[c].l,l,m);
else return query(pos,p[c].r,m+1,r);
}
int main()
{
scanf("%d",&n);
memset(la,-1,sizeof la);
for (int i = 1; i <= n; ++i)
scanf("%d",a+i);
root[0] = build(1,n);
for (int i = 1 ; i <= n; ++i)
{
int v = a[i];
if (la[v] == -1) root[i] = update(i,root[i-1],1,1,n);
else{
int t = update(la[v],root[i-1],-1,1,n);
root[i] = update(i,t,1,1,n);
}
la[v] = i;
}
scanf("%d",&q);
while(q--)
{
int x,y;
scanf("%d %d",&x, &y);
printf("%d
",query(x,root[y],1,n));
}
return 0;
}
平衡树
fhq_treap
-
插入x数
-
删除x数(若有多个相同的数，因只删除一个)
-
查询x数的排名(若有多个相同的数，因输出最小的排名)
-
查询排名为x的数
-
求x的前驱(前驱定义为小于x，且最大的数)
-
求x的后继(后继定义为大于x，且最小的数)
const int maxn=1e5+5;
struct node
{
int son[2],v,rnd,size;
}tr[maxn];
int n,m,l,r;
int tot,root;
int new_node(int v)//创建权值为v的结点。
{
tot++;
tr[tot].size=1;
tr[tot].v=v;
tr[tot].rnd=rand();
tr[tot].son[0]=tr[tot].son[1]=0;
return tot;
}
void update(int x)
{
tr[x].size=tr[tr[x].son[0]].size+tr[tr[x].son[1]].size+1;
}
int merge(int x,int y)
{
if(!x||!y)
return x+y;
if(tr[x].rnd<tr[y].rnd)
{
tr[x].son[1]=merge(tr[x].son[1],y);
update(x);
return x;
}
else
{
tr[y].son[0]=merge(x,tr[y].son[0]);
update(y);
return y;
}
}
void split(int now,int k,int &x,int &y)//以权值k分离now树成x,y
{
if(!now) x=y=0;
else
{
if(tr[now].v<=k) //把所有小于等于k的权值的节点分到一棵树中
x=now,split(tr[now].son[1],k,tr[now].son[1],y);
else
y=now,split(tr[now].son[0],k,x,tr[now].son[0]);
update(now);
}
}
void rev(int l,int r)
{
int x,y,u,v;
split(root,r+1,x,y);
split(x,l,u,v);
root=merge(merge(u,v),y);
}
void insert(int v)
{
int x,y;
split(root,v,x,y);
root=merge(merge(x,new_node(v)),y);
}
void del(int v)
{
int x,y,z;
split(root,v,x,z);
split(x,v-1,x,y);
y=merge(tr[y].son[0],tr[y].son[1]);
root=merge(merge(x,y),z);
}
void findrank(int v)
{
int x,y;
split(root,v-1,x,y);
printf("%d
",tr[x].size+1);
root=merge(x,y);
}
int kth(int now,int k)
{
while(1)
{
if(k<=tr[tr[now].son[0]].size)
now=tr[now].son[0];
else
{
if(k==tr[tr[now].son[0]].size+1) return now;
else
{
k-=tr[tr[now].son[0]].size+1;
now=tr[now].son[1];
}
}
}
}
void precursor(int v)
{
int x,y;
split(root,v-1,x,y);
printf("%d
",tr[kth(x,tr[x].size)].v);
root=merge(x,y);
}
void successor(int v)
{
int x,y;
split(root,v,x,y);
printf("%d
",tr[kth(y,1)].v);
root=merge(x,y);
}
int main()
{
srand((unsigned)time(NULL));
int n,op,v;
scanf("%d",&n);
while(n--)
{
scanf("%d%d",&op,&v);
switch(op)
{
case 1:insert(v);break;
case 2:del(v);break;
case 3:findrank(v);break;
case 4:printf("%d
",tr[kth(root,v)].v);break;
case 5:precursor(v);break;
case 6:successor(v);break;
default:break;
}
}
return 0;
}
可持久化treap
1.插入x数
2.删除x数(若有多个相同的数，因只删除一个，如果没有请忽略该操作)
3.查询x数的排名(排名定义为比当前数小的数的个数+1。若有多个相同的数，因输出最小的排名)
4.查询排名为x的数
5.求x的前驱(前驱定义为小于x，且最大的数，如不存在输出-2147483647)
6.求x的后继(后继定义为大于x，且最小的数，如不存在输出2147483647)
第 i 行记为 ，表示基于的过去版本号
#define SF scanf
#define PF printf
#define MAXN 500010
using namespace std;
struct node{
int ch[2];
int key,sum,fix;
node () {}
node (int key1):key(key1),fix(rand()),sum(1) {}
}Tree[MAXN*30];
int ncnt,root[MAXN];
void update(int x){
Tree[x].sum=Tree[Tree[x].ch[0]].sum+Tree[Tree[x].ch[1]].sum+1;
}
int Merge(int x,int y){
if(x==0) return y;
if(y==0) return x;
if(Tree[x].fix<Tree[y].fix){
Tree[x].ch[1]=Merge(Tree[x].ch[1],y);
update(x);
return x;
}
else{
Tree[y].ch[0]=Merge(x,Tree[y].ch[0]);
update(y);
return y;
}
}
pair<int,int> Split(int x,int k){
if(x==0)
return make_pair(0,0);
int newone;
pair<int,int> y;
if(Tree[x].key<=k){
newone=++ncnt;
Tree[newone]=Tree[x];
y=Split(Tree[newone].ch[1],k);
Tree[newone].ch[1]=y.first;
update(newone);
y.first=newone;
}
else{
newone=++ncnt;
Tree[newone]=Tree[x];
y=Split(Tree[newone].ch[0],k);
Tree[newone].ch[0]=y.second;
update(newone);
y.second=newone;
}
return y;
}
int find_kth(int now,int val){
pair<int,int> x=Split(root[now],val-1);
int ans=Tree[x.first].sum+1;
root[now]=Merge(x.first,x.second);
return ans;
}
int get_kth(int x,int k){
if(x==0)
return 0;
int ch0=Tree[x].ch[0];
if(Tree[ch0].sum>=k)
return get_kth(ch0,k);
if(Tree[ch0].sum+1==k)
return Tree[x].key;
return get_kth(Tree[x].ch[1],k-Tree[ch0].sum-1);
}
int get_pre(int now,int val){
int k=find_kth(now,val);
return get_kth(root[now],k-1);
}
int get_bac(int now,int val){
pair<int,int> x=Split(root[now],val);
int ans=get_kth(x.second,1);
root[now]=Merge(x.first,x.second);
return ans;
}
void Insert(int val,int now){
pair<int,int> x=Split(root[now],val);
Tree[++ncnt]=node(val);
Tree[ncnt].ch[0]=0;
Tree[ncnt].ch[1]=0;
root[now]=Merge(Merge(x.first,ncnt),x.second);
}
void Delete(int val,int now){
pair<int,int> x=Split(root[now],val);
pair<int,int> y=Split(x.first,val-1);
if(Tree[y.second].key==val)
y.second=Merge(Tree[y.second].ch[0],Tree[y.second].ch[1]);
root[now]=Merge(Merge(y.first,y.second),x.second);
}
int n,las,x,tag;
int main(){
SF("%d",&n);
Insert(-2147483647,0);
Insert(2147483647,0);
for(int i=1;i<=n;i++){
SF("%d%d%d",&las,&tag,&x);
if(Tree[root[las]].sum!=0){
root[i]=++ncnt;
Tree[root[i]]=Tree[root[las]];
}
if(tag==1)
Insert(x,i);
if(tag==2)
Delete(x,i);
if(tag==3)
PF("%d
",find_kth(i,x)-1);
if(tag==4)
PF("%d
",get_kth(root[i],x+1));
if(tag==5)
PF("%d
",get_pre(i,x));
if(tag==6)
PF("%d
",get_bac(i,x));
}
}
Segment Beats
维护一个序列，支持下面几种操作：
1.给一个区间[L, R]加上一个数x。
2.把一个区间[L, R]里小于x的数变成x。
3.把一个区间[L, R]里大于x的数变成x。
4.求区间[L, R]的和。
5.求区间[L, R]的最大值。
6.求区间[L, R]的最小值。
const int N = 1050000, inf = ~0U >> 1;
int n, m, i, a[500010], op, c, d, p, ans;
int len[N], cma[N], cmi[N], ma[N], ma2[N], mi[N], mi2[N], tma[N], tmi[N], ta[N];
long long sum[N], ret;
void tagma(int x, int p)
{
tma[x] += p;
if (ma[x] == mi[x])
{
ma[x] += p;
mi[x] = ma[x];
sum[x] = 1LL * ma[x] * len[x];
return;
}
ma[x] += p;
if (cma[x] + cmi[x] == len[x])
mi2[x] += p;
sum[x] += 1LL * p * cma[x];
}
void tagmi(int x, int p)
{
tmi[x] += p;
if (ma[x] == mi[x])
{
ma[x] += p;
mi[x] = ma[x];
sum[x] = 1LL * ma[x] * len[x];
return;
}
mi[x] += p;
if (cma[x] + cmi[x] == len[x])
ma2[x] += p;
sum[x] += 1LL * p * cmi[x];
}
void taga(int x, int p)
{
ta[x] += p;
ma[x] += p;
mi[x] += p;
if (ma2[x] !=-inf)
ma2[x] += p;
if (mi2[x] != inf)
mi2[x] += p;
sum[x] += 1LL * p * len[x];
}
void pb(int x)
{
if (tma[x])
{
if (ma[x << 1] > ma[x << 1 | 1])
tagma(x << 1, tma[x]);
else if (ma[x << 1] < ma[x << 1 | 1])
tagma(x << 1 | 1, tma[x]);
else
{
tagma(x << 1, tma[x]);
tagma(x << 1 | 1, tma[x]);
}
tma[x] = 0;
}
if (tmi[x])
{
if (mi[x << 1] < mi[x << 1 | 1])
tagmi(x << 1, tmi[x]);
else if (mi[x << 1] > mi[x << 1 | 1])
tagmi(x << 1 | 1, tmi[x]);
else
{
tagmi(x << 1, tmi[x]);
tagmi(x << 1 | 1, tmi[x]);
}
tmi[x] = 0;
}
if (ta[x])
{
taga(x << 1, ta[x]);
taga(x << 1 | 1, ta[x]);
ta[x] = 0;
}
}
void up(int x)
{
sum[x] = sum[x << 1] + sum[x << 1 | 1];
if (ma[x << 1] > ma[x << 1 | 1])
{
ma[x] = ma[x << 1];
ma2[x] = max(ma2[x << 1], ma[x << 1 | 1]);
cma[x] = cma[x << 1];
}
else if (ma[x << 1] < ma[x << 1 | 1])
{
ma[x] = ma[x << 1 | 1];
ma2[x] = max(ma[x << 1], ma2[x << 1 | 1]);
cma[x] = cma[x << 1 | 1];
}
else
{
ma[x] = ma[x << 1];
ma2[x] = max(ma2[x << 1], ma2[x << 1 | 1]);
cma[x] = cma[x << 1] + cma[x << 1 | 1];
}
if (mi[x << 1] < mi[x << 1 | 1])
{
mi[x] = mi[x << 1];
mi2[x] = min(mi2[x << 1], mi[x << 1 | 1]);
cmi[x] = cmi[x << 1];
}
else if (mi[x << 1] > mi[x << 1 | 1])
{
mi[x] = mi[x << 1 | 1];
mi2[x] = min(mi[x << 1], mi2[x << 1 | 1]);
cmi[x] = cmi[x << 1 | 1];
}
else
{
mi[x] = mi[x << 1];
mi2[x] = min(mi2[x << 1], mi2[x << 1 | 1]);
cmi[x] = cmi[x << 1] + cmi[x << 1 | 1];
}
}
void build(int x, int a, int b)
{
len[x] = b-a + 1;
if (a == b)
{
ma[x] = mi[x] = sum[x] = ::a[a], ma2[x] =-inf, mi2[x] = inf;
cma[x] = cmi[x] = 1;
return;
}
int mid = (a + b) >> 1;
build(x << 1, a, mid), build(x << 1 | 1, mid + 1, b);
up(x);
}
void change(int x, int a, int b)
{
if (c <= a && b <= d)
{
taga(x, p);
return;
}
pb(x);
int mid = (a + b) >> 1;
if (c <= mid)
change(x << 1, a, mid);
if (d > mid)
change(x << 1 | 1, mid + 1, b);
up(x);
}
void cmax(int x, int a, int b)
{
if (c <= a && b <= d)
{
if (mi[x] >= p)
return;
if (mi2[x] > p)
{
tagmi(x, p-mi[x]);
return;
}
}
pb(x);
int mid = (a + b) >> 1;
if (c <= mid)
cmax(x << 1, a, mid);
if (d > mid)
cmax(x << 1 | 1, mid + 1, b);
up(x);
}
void cmin(int x, int a, int b)
{
if (c <= a && b <= d)
{
if (ma[x] <= p)
return;
if (ma2[x] < p)
{
tagma(x, p-ma[x]);
return;
}
}
pb(x);
int mid = (a + b) >> 1;
if (c <= mid)
cmin(x << 1, a, mid);
if (d > mid)
cmin(x << 1 | 1, mid + 1, b);
up(x);
}
void qsum(int x, int a, int b)
{
if (c <= a && b <= d)
{
ret += sum[x];
return;
}
pb(x);
int mid = (a + b) >> 1;
if (c <= mid)
qsum(x << 1, a, mid);
if (d > mid)
qsum(x << 1 | 1, mid + 1, b);
}
void qmax(int x, int a, int b)
{
if (c <= a && b <= d)
{
ans = max(ans, ma[x]);
return;
}
pb(x);
int mid = (a + b) >> 1;
if (c <= mid)
qmax(x << 1, a, mid);
if (d > mid)
qmax(x << 1 | 1, mid + 1, b);
}
void qmin(int x, int a, int b)
{
if (c <= a && b <= d)
{
ans = min(ans, mi[x]);
return;
}
pb(x);
int mid = (a + b) >> 1;
if (c <= mid)
qmin(x << 1, a, mid);
if (d > mid)
qmin(x << 1 | 1, mid + 1, b);
}
int main()
{
for (read(n), i = 1; i <= n; i++)
read(a[i]);
build(1, 1, n);
read(m);
while (m--)
{
read(op), read(c), read(d);
if (op == 1)
read(p), change(1, 1, n);
if (op == 2)
read(p), cmax(1, 1, n);
if (op == 3)
read(p), cmin(1, 1, n);
if (op == 4)
ret = 0, qsum(1, 1, n), printf("% lld
", ret);
if (op == 5)
ans =-inf, qmax(1, 1, n), printf("% d
", ans);
if (op == 6)
ans = inf, qmin(1, 1, n), printf("% d
", ans);
}
}
动态树
树的连通性
Connect u v：链接uv；
Destroy u v：断开uv；
Query u v：询问uv是否联通。
int n,m;
struct data
{
int fa,lc,rc,rev,pfa;
}a[10010];
void Pushdown(int x)
{
if (a[x].rev)
{
a[x].rev=0;
a[a[x].lc].rev^=1;
a[a[x].rc].rev^=1;
swap(a[x].lc,a[x].rc);
}
}
void Pushpath(int x)
{
if (a[x].fa!=0) Pushpath(a[x].fa);
Pushdown(x);
}
void zag(int x)
{
int y=a[x].fa;
a[x].pfa=a[y].pfa;a[y].pfa=0;
a[y].rc=a[x].lc;
if (a[x].lc!=0) a[a[x].lc].fa=y;
a[x].fa=a[y].fa;
if (a[y].fa!=0)
{
if (y==a[a[y].fa].lc) a[a[y].fa].lc=x;
else a[a[y].fa].rc=x;
}
a[y].fa=x;a[x].lc=y;
}
void zig(int x)
{
int y=a[x].fa;
a[x].pfa=a[y].pfa;a[y].pfa=0;
a[y].lc=a[x].rc;
if (a[x].rc!=0) a[a[x].rc].fa=y;
a[x].fa=a[y].fa;
if (a[y].fa!=0)
{
if (y==a[a[y].fa].lc) a[a[y].fa].lc=x;
else a[a[y].fa].rc=x;
}
a[y].fa=x;a[x].rc=y;
}
void spaly(int x)
{
int y,z;
Pushpath(x);
while (a[x].fa!=0)
{
y=a[x].fa;
if (a[y].fa==0)
{
if (x==a[y].lc) zig(x);else zag(x);
}
else if (a[y].lc==x)
{
z=a[y].fa;
if (y==a[z].lc) zig(y),zig(x);else zig(x),zag(x);
}
else
{
z=a[y].fa;
if (y==a[z].lc) zag(x),zig(x);else zag(y),zag(x);
}
}
}
void Access(int x)
{
spaly(x);Pushdown(x);
int y=a[x].rc;
a[x].rc=a[y].fa=0;
a[y].pfa=x;y=a[x].pfa;
while (y!=0)
{
spaly(y);
Pushdown(y);
a[a[y].rc].fa=0;
a[a[y].rc].pfa=y;
a[y].rc=x;
a[x].fa=y;a[x].pfa=0;
x=y;
y=a[x].pfa;
}
spaly(x);
}
void Makeroot(int x)
{
Access(x);
spaly(x);
a[x].rev^=1;
Pushdown(x);
}
int Findroot(int x)
{
Access(x);
spaly(x);
Pushdown(x);
while (a[x].lc!=0) x=a[x].lc,Pushdown(x);
spaly(x);
return x;
}
void Join(int x,int y)
{
Makeroot(x);
spaly(x);
Pushdown(x);
Access(y);
spaly(y);
Pushdown(y);
a[x].lc=y;a[y].fa=x;
}
void Cut(int x,int y)
{
Makeroot(x);
Access(y);
spaly(y);
Pushdown(y);
a[a[y].lc].fa=0;
a[y].lc=0;
}
int main()
{
freopen("1.in","r",stdin);
freopen("1.out","w",stdout);
read(n);read(m);
for (int i=1;i<=n;i++)
a[i].fa=a[i].lc=a[i].rc=a[i].rev=a[i].pfa=0;
int x,y;char order[20];
while (m--)
{
x=read(order);
read(x);read(y);
if (order[0]=='Q')
{
if (Findroot(x)==Findroot(y)) puts("Yes");
else puts("No");
}
else if (order[0]=='C') Join(x,y);
else if (order[0]=='D') Cut(x,y);
}
return 0;
}
链和与最大最小
CHANGE u t : 把结点u的权值改为t
QMAX u v: 询问从点u到点v的路径上的节点的最大权值
QSUM u v: 询问从点u到点v的路径上的节点的权值和
int n,m;
struct data
{
int fa,lc,rc,rev,pfa;
LL xsum,lsum,rsum,lmax,rmax,xmax;
}a[300010];
void Pushdown(int x)
{
if (a[x].rev)
{
a[x].rev=0;
a[a[x].lc].rev^=1;
a[a[x].rc].rev^=1;
swap(a[x].lc,a[x].rc);
swap(a[x].lsum,a[x].rsum);
swap(a[x].lmax,a[x].rmax);
}
}
void Pushpath(int x)
{
if (a[x].fa!=0) Pushpath(a[x].fa);
Pushdown(x);
}
void zag(int x)
{
int y=a[x].fa;
a[x].pfa=a[y].pfa;a[y].pfa=0;
a[y].rc=a[x].lc;
if (a[x].lc!=0) a[a[x].lc].fa=y;
a[x].fa=a[y].fa;
if (a[y].fa!=0)
{
if (y==a[a[y].fa].lc) a[a[y].fa].lc=x;
else a[a[y].fa].rc=x;
}
a[y].fa=x;a[x].lc=y;
int t=a[x].lsum;a[x].lsum+=a[y].xsum+a[y].lsum;a[y].rsum=t;
LL q=a[x].lmax;a[x].lmax=max(max(a[x].lmax,a[y].xmax),a[y].lmax);a[y].rmax=q;
}
void zig(int x)
{
int y=a[x].fa;
a[x].pfa=a[y].pfa;a[y].pfa=0;
a[y].lc=a[x].rc;
if (a[x].rc!=0) a[a[x].rc].fa=y;
a[x].fa=a[y].fa;
if (a[y].fa!=0)
{
if (y==a[a[y].fa].lc) a[a[y].fa].lc=x;
else a[a[y].fa].rc=x;
}
a[y].fa=x;a[x].rc=y;
int t=a[x].rsum;a[x].rsum+=a[y].xsum+a[y].rsum;a[y].lsum=t;
LL q=a[x].rmax;a[x].rmax=max(max(a[x].rmax,a[y].xmax),a[y].rmax);a[y].lmax=q;
}
void spaly(int x)
{
int y,z;
Pushpath(x);
while (a[x].fa!=0)
{
y=a[x].fa;
if (a[y].fa==0)
{
if (x==a[y].lc) zig(x);else zag(x);
}
else if (a[y].lc==x)
{
z=a[y].fa;
if (y==a[z].lc) zig(y),zig(x);else zig(x),zag(x);
}
else
{
z=a[y].fa;
if (y==a[z].lc) zag(x),zig(x);else zag(y),zag(x);
}
}
}
void Access(int x)
{
spaly(x);Pushdown(x);
int y=a[x].rc;
a[x].rc=a[y].fa=0;
a[x].rsum=0;a[x].rmax=-30010LL*200010LL;
a[y].pfa=x;y=a[x].pfa;
while (y!=0)
{
spaly(y);
Pushdown(y);
a[a[y].rc].fa=0;
a[a[y].rc].pfa=y;
a[y].rc=x;
a[y].rsum=a[x].lsum+a[x].rsum+a[x].xsum;
a[y].rmax=max(max(a[x].lmax,a[x].rmax),a[x].xmax);
a[x].fa=y;a[x].pfa=0;
x=y;
y=a[x].pfa;
}
spaly(x);
}
void Makeroot(int x)
{
Access(x);
spaly(x);
a[x].rev^=1;
Pushdown(x);
}
int Findroot(int x)
{
Access(x);
spaly(x);
Pushdown(x);
while (a[x].lc!=0) x=a[x].lc,Pushdown(x);
spaly(x);
return x;
}
void Join(int x,int y)
{
Makeroot(x);
spaly(x);
Pushdown(x);
Access(y);
spaly(y);
Pushdown(y);
a[x].lc=y;a[y].fa=x;
a[x].lsum=a[y].xsum+a[y].lsum+a[y].rsum;
a[x].lmax=max(max(a[y].xmax,a[y].lmax),a[y].rmax);
}
void Cut(int x,int y)
{
Makeroot(x);
Access(y);
spaly(y);
Pushdown(y);
a[a[y].lc].fa=0;
a[y].lc=0;
a[y].lsum=0;a[y].lmax=-30010LL*200010LL;
}
void Updata(int x,int y)
{
spaly(x);
Pushdown(x);
a[x].xmax=y;a[x].xsum=y;
}
LL querymax(int x,int y)
{
Makeroot(x);
Access(y);
spaly(y);
Pushdown(y);
return max(a[y].lmax,a[y].xmax);
}
LL querysum(int x,int y)
{
Makeroot(x);
Access(y);
spaly(y);
Pushdown(y);
return a[y].lsum+a[y].xsum;
}
int main()
{
freopen("bzoj_1036.in","r",stdin);
freopen("bzoj_1036.out","w",stdout);
read(n);
memset(a,0,sizeof(a));
for (int i=1;i<=n;i++)
a[i].lmax=-30010LL*200010LL,a[i].rmax=-30010LL*200010LL,a[i].xmax=-30010LL*200010LL;
int x,y;LL z;char o[100];
for (int i=1;i<n;i++)
read(x),read(y),Join(x,y);
for (int i=1;i<=n;i++)
read(z),Updata(i,z);
read(m);
while (m--)
{
x=read(o);
if (o[1]=='M') read(x),read(y),print(querymax(x,y)),puts("");
else if (o[1]=='S') read(x),read(y),print(querysum(x,y)),puts("");
else read(x),read(z),Updata(x,z);
}
return 0;
}
树链剖分
子树和与单链和
int n,m,point[100010],po,T;
struct data
{
int fa,top,c,ne,d,point,l,r;
LL x;
}a[100010];
vector<int> b[100010];
bool v[100010];
struct ST
{
LL x,tag;
}tree[400010];
void dfs1(int x)
{
v[x]=true;
int Max=0,w=0;
for (int i=0;i<b[x].size();i++)
if (!v[b[x][i]])
{
a[b[x][i]].d=a[x].d+1;
dfs1(b[x][i]);
if (a[b[x][i]].c>Max) Max=a[b[x][i]].c,w=b[x][i];
a[x].c+=a[b[x][i]].c;
a[b[x][i]].fa=x;
}
a[x].ne=w;a[x].c++;
}
void dfs2(int x,int y)
{
v[x]=true,point[++po]=x,a[x].point=po,a[x].top=y;a[x].l=po;
if (!v[a[x].ne] && a[x].ne!=0) dfs2(a[x].ne,a[x].top);
for (int i=0;i<b[x].size();i++)
if (!v[b[x][i]])
dfs2(b[x][i],b[x][i]);
a[x].r=po;
}
void Pushdown(int x,int l,int r)
{
if (tree[x].tag!=0)
{
tree[x<<1].tag+=tree[x].tag;
tree[x<<1|1].tag+=tree[x].tag;
tree[x].x+=tree[x].tag*(LL)(r-l+1);
tree[x].tag=0;
}
}
void change(int lq,int rq,int x,int l,int r,LL z)
{
if (lq<=l && rq>=r) {tree[x].tag+=z;return ;}
int mid=l+r>>1;
Pushdown(x,l,r);
if (lq<=mid) change(lq,rq,x<<1,l,mid,z);
if (rq>mid) change(lq,rq,x<<1|1,mid+1,r,z);
tree[x].x=tree[x<<1].x+tree[x<<1|1].x+tree[x<<1].tag*(LL)(mid-l+1)+tree[x<<1|1].tag*(LL)(r-mid);
}
LL query(int lq,int rq,int x,int l,int r)
{
if (lq<=l && rq>=r) return tree[x].x+tree[x].tag*(LL)(r-l+1);
int mid=l+r>>1;
LL res=0;
Pushdown(x,l,r);
if (lq<=mid) res+=query(lq,rq,x<<1,l,mid);
if (rq>mid) res+=query(lq,rq,x<<1|1,mid+1,r);
tree[x].x=tree[x<<1].x+tree[x<<1|1].x+tree[x<<1].tag*(LL)(mid-l+1)+tree[x<<1|1].tag*(LL)(r-mid);
return res;
}
void build(int l,int r,int x)
{
if (l==r) {tree[x].x=a[point[l]].x;return ;}
int mid=l+r>>1;
build(l,mid,x<<1);build(mid+1,r,x<<1|1);
tree[x].x=tree[x<<1].x+tree[x<<1|1].x;
tree[x].tag=0;
}
int main()
{
read(n);read(m);
int x,y;LL z;
memset(a,0,sizeof(a));
for (int i=1;i<=n;i++)
read(a[i].x);
for (int i=1;i<n;i++)
read(x),read(y),b[x].push_back(y),b[y].push_back(x);
memset(tree,0,sizeof(tree));
memset(v,false,sizeof(v));
a[1].d=1;a[1].fa=1;
dfs1(1);
memset(v,false,sizeof(v));
x=1;int s=0;po=0;
dfs2(1,1);
build(1,n,1);
while(m--)
{
read(x);
if (x==1) read(x),read(z),change(a[x].point,a[x].point,1,1,n,z);
else if (x==2) read(x),read(z),change(a[x].l,a[x].r,1,1,n,z);
else if (x==3)
{
read(x);LL res=0;
while (a[x].top!=1)
{
res+=query(a[a[x].top].point,a[x].point,1,1,n);
x=a[a[x].top].fa;
}
res+=query(a[1].point,a[x].point,1,1,n);
print(res);putchar('
');
}
}
return 0;
}
LCA
int n,m,point[50010],po,T,way[50010],bu[50010];
struct data
{
int fa,top,c,ne,d,point,l,r;
LL x;
LL b[250];
}a[50010];
struct dara
{
int x;LL y;
dara(int a,LL b):x(a),y(b){};
};
vector<dara> b[50010];
bool v[50010];
void dfs1(int x)
{
v[x]=true;
int Max=0,w=0;
for (int i=0;i<b[x].size();i++)
if (!v[b[x][i].x])
{
a[b[x][i].x].d=a[x].d+1;
a[b[x][i].x].x=b[x][i].y;
dfs1(b[x][i].x);
if (a[b[x][i].x].c>Max) Max=a[b[x][i].x].c,w=b[x][i].x;
a[x].c+=a[b[x][i].x].c;
a[b[x][i].x].fa=x;
}
a[x].ne=w;a[x].c++;
}
void dfs2(int x,int y)
{
v[x]=true,point[++po]=x,a[x].point=po,a[x].top=y;a[x].l=po;
if (!v[a[x].ne] && a[x].ne!=0) dfs2(a[x].ne,a[x].top);
for (int i=0;i<b[x].size();i++)
if (!v[b[x][i].x])
dfs2(b[x][i].x,b[x][i].x);
a[x].r=po;
}
void calc(int x,int y)
{
int z=x;v[x]=true;
for (int i=1;i<=y;i++)
{
z=a[z].fa;
a[x].b[i]=a[z].b[i]+a[x].x;
}
for (int i=0;i<b[x].size();i++)
if (!v[b[x][i].x]) calc(b[x][i].x,y);
}
int Find(int x,int y) //找x向上y个
{
while (a[x].d-a[a[x].top].d<y) y-=a[x].d-a[a[x].top].d+1,x=a[a[x].top].fa;
return point[a[x].point-y];
}
int lca(int x,int y) //点x、y的lca
{
while (a[x].top!=a[y].top)
{
if (a[a[x].top].d<a[a[y].top].d) swap(x,y);
x=a[a[x].top].fa;
}
if (a[x].d>a[y].d) swap(x,y);
return x;
}
int dist(int x,int y) //点x、y距离
{
int z=lca(x,y);
return a[x].d+a[y].d-2*a[z].d;
}
int Ans(int x,int y)
{
LL b[50010];
int s=0;
while (x!=y)
{
if (a[x].d<a[y].d) swap(x,y);
b[++s]=a[x].x;
x=a[x].fa;
}
sort(b+1,b+1+s);
return b[(s+1)/2];
}
int main()
{
freopen("draw.in","r",stdin);
freopen("draw.out","w",stdout);
read(n);
int x,y;LL z;
memset(a,0,sizeof(a));
for (int i=1;i<n;i++)
read(x),read(y),read(z),b[x].push_back(dara(y,z)),b[y].push_back(dara(x,z));
memset(v,false,sizeof(v));
a[1].d=1;a[1].fa=1;
dfs1(1);
memset(v,false,sizeof(v));
x=1;int s=0;po=0;
dfs2(1,1);
LL ans=0;
for (int i=1;i<n;i++)
for (int j=i+1;j<=n;j++)
if (dist(i,j)%2==1) ans+=Ans(i,j);
print(ans);puts("");
return 0;
}
可并堆
Merger(i, j)。把i所在的团和j所在的团合并成一个团。如果i, j有一个人是死人，那么就忽略该命令。
Kill(i)。把i所在的团里面得分最低的人杀死。如果i这个人已经死了，这条命令就忽略。 皇帝希望他每发布一条kill命令，下面的将军就把被杀的人的分数报上来。
int n,m,fa[1000010];
bool v[1000010];
struct data
{
int key,dist,l,r;
}tree[1000010];
int merge(int x,int y)
{
if (x==0 || y==0) return x+y;
if (tree[x].key>tree[y].key) swap(x,y);
tree[x].r=merge(tree[x].r,y);
if (tree[tree[x].r].dist>tree[tree[x].l].dist) swap(tree[x].l,tree[x].r);
if (tree[x].r==0) tree[x].dist=0;
else tree[x].dist=tree[tree[x].r].dist+1;
return x;
}
int find(int x)
{
return x==fa[x]?x:fa[x]=find(fa[x]);
}
int main()
{
scanf("%d",&n);
for (int i=1;i<=n;i++)
scanf("%d",&tree[i].key),fa[i]=i;
scanf("%d",&m);
char x[10];int y,z;
memset(v,false,sizeof(v));
while (m--)
{
scanf("%s",x);
if (x[0]=='M')
{
scanf("%d%d",&y,&z);
if (v[y] || v[z]) continue;
y=find(y);z=find(z);
if (y==z) continue;
fa[z]=fa[y]=merge(y,z);
}
else
{
scanf("%d",&y);z=find(y);
if (v[y]) {printf("0
");continue;}else v[z]=true;
printf("%d
",tree[z].key);
int p=merge(tree[fa[y]].l,tree[fa[y]].r);
fa[fa[y]]=p;fa[p]=p;
}
}
return 0;
}
CDQ
三维偏序（计数）
满足且且且的数对的个数
typedef long long ll;
const int N=5e4+10;
struct Node { int b,c,d,e; } a[N], t1[N], t2[N];
ll ans;
int C[N];
#define lowbit(x) x&-x
void updata(int x, int d) {
while(x<N) {
C[x]+=d;
x+=lowbit(x);
}
}
int query(int x) {
int ret=0;
while(x) {
ret+=C[x];
x-=lowbit(x);
}
return ret;
}
void merge2(int l, int r) {
if(l==r) return;
int mid=(l+r)>>1;
merge2(l, mid), merge2(mid+1, r);
int i=l, j=mid+1, k=l;
Node *a=t1, *t=t2;
while(i<=mid && j<=r) {
if(a[i].c<a[j].c) {
if(!a[i].e) updata(a[i].d, 1);
t[k++]=a[i++];
}
else {
if(a[j].e) ans+=query(a[j].d);
t[k++]=a[j++];
}
}
while(j<=r) {
if(a[j].e) ans+=query(a[j].d);
t[k++]=a[j++];
}
for(int ni=l;ni<i;ni++) if(!a[ni].e) updata(a[ni].d, -1);
while(i<=mid) t[k++]=a[i++];
for(int i=l;i<=r;i++) a[i]=t[i];
}
void merge(int l, int r) {
if(l==r) return;
int mid=(l+r)>>1;
merge(l, mid), merge(mid+1, r);
Node *t=t1;
int i=l, j=mid+1, k=l;
while(i<=mid && j<=r) {
if(a[i].b<a[j].b) a[i].e=0, t[k++]=a[i++];
else a[j].e=1, t[k++]=a[j++];
}
while(i<=mid) a[i].e=0, t[k++]=a[i++];
while(j<=r) a[j].e=1, t[k++]=a[j++];
for(int i=l;i<=r;i++) a[i]=t[i];
merge2(l, r);
}
template<typename T> void gn(T &x) {
x=0;
char ch=getchar();
while(ch<'0' || ch>'9') ch=getchar();
while(ch>='0' && ch<='9') x=x*10+ch-'0', ch=getchar();
}
int main() {
int n;
gn(n);
for(int i=1;i<=n;i++) gn(a[i].b);
for(int i=1;i<=n;i++) gn(a[i].c);
for(int i=1;i<=n;i++) gn(a[i].d);
merge(1, n);
printf("%lld
", ans);
return 0;
}
四维偏序（计数）
满足且且且且的数对的个数
#define Inf 2e9
#define Lowbit(x) (x&-x)
const int maxn=55000;
struct Node{
int id,x,y,z,q,flag1,flag2;
}a[maxn],b[maxn],c[maxn],d[maxn];
int ans=0,C[maxn],tim[maxn],Tim,n;
int Query(int x){
int res=0;
for(int i=x;i;i-=Lowbit(i)){
if(tim[i]!=Tim){tim[i]=Tim;C[i]=0;}
res+=C[i];
}
return res;
}
void Update(int x){
for(int i=x;i<=n;i+=Lowbit(i)){
if(tim[i]!=Tim){tim[i]=Tim;C[i]=0;}
C[i]++;
}
}
void CDQ3(int l,int r){
if(l>=r)return;
int mid=(l+r)>>1;
CDQ3(l,mid);CDQ3(mid+1,r);
Tim++;
int i=l,j=mid+1,k=l;
while(i<=mid && j<=r){
if(c[i].z<c[j].z){
if(c[i].flag1 && c[i].flag2)Update(c[i].q);
d[k++]=c[i++];
}
else {
if(!c[j].flag1 && !c[j].flag2)ans+=Query(c[j].q);
d[k++]=c[j++];
}
}
while(i<=mid){
if(c[i].flag1 && c[i].flag2)Update(c[i].q);
d[k++]=c[i++];
}
while(j<=r){
if(!c[j].flag1 && !c[j].flag2)ans+=Query(c[j].q);
d[k++]=c[j++];
}
for(i=l;i<=r;i++)c[i]=d[i];
}
void CDQ2(int l,int r){
if(l>=r)return;
int mid=(l+r)>>1;
CDQ2(l,mid);CDQ2(mid+1,r);
int i=l,j=mid+1,k=l;
while(i<=mid && j<=r){
if(b[i].y<b[j].y){
c[k].flag2=b[i].flag2=true;c[k++]=b[i++];
}
else {
c[k].flag2=b[j].flag2=false;c[k++]=b[j++];
}
}
while(i<=mid){
c[k].flag2=b[i].flag2=true;c[k++]=b[i++];
}
while(j<=r){
c[k].flag2=b[j].flag2=false;c[k++]=b[j++];
}
for(i=l;i<=r;i++)b[i]=c[i];
CDQ3(l,r);
}
void CDQ1(int l,int r){
if(l>=r)return;
int mid=(l+r)>>1;
CDQ1(l,mid);CDQ1(mid+1,r);
int i=l,j=mid+1,k=l;
while(i<=mid && j<=r){
if(a[i].x<a[j].x)b[k++]=a[i++];
else b[k++]=a[j++];
}
while(i<=mid)b[k++]=a[i++];
while(j<=r)b[k++]=a[j++];
for(i=l;i<=r;i++){
a[i]=b[i];
if(a[i].id<=mid)a[i].flag1=b[i].flag1=true;
else a[i].flag1=b[i].flag1=false;
}
CDQ2(l,r);
}
void Init();
int main(){
freopen("partial_order_two.in","r",stdin);freopen("partial_order_two.out","w",stdout);
Init();
return 0;
}
void Init(){
scanf("%d",&n);
for(int i=1;i<=n;i++)a[i].id=i,scanf("%d",&a[i].x);
for(int i=1;i<=n;i++)scanf("%d",&a[i].y);
for(int i=1;i<=n;i++)scanf("%d",&a[i].z);
for(int i=1;i<=n;i++)scanf("%d",&a[i].q);
CDQ1(1,n);
printf("%d
",ans);
}
k维偏序（计数）
求的数量。
typedef bitset<40001>bit;
const int N=40005;
int n,k,siz,block[N];
struct Bitset{
int a[N],lis[N];
//lis[N]表示权值为n的点的编号
bit B[201];
//bit内储存第i维权值小于j*j的向量
bit get(int p){
p=a[p]; int bp=block[p]; bit ans=B[bp-1];
for(int i=(bp-1)*siz+1;i<p;i++) ans.set(lis[i]);
return ans;
}
}d[7];
void build(int op){
for(int i=1;i<=n;i++) d[op].lis[d[op].a[i]]=i;
bit t; t.reset();
for(int i=1;i<=n;i++){
t.set(d[op].lis[i]);
if(i%siz==0) d[op].B[i/siz]=t;
}
}
int main(){
freopen("partial_order_plus.in","r",stdin);
freopen("partial_order_plus.out","w",stdout);
scanf("%d%d",&n,&k); siz=int(sqrt(n*1.0));
for(int i=1;i<=n;i++) block[i]=(i-1)/siz+1;
for(int i=1;i<=n;i++) d[0].a[i]=i;
for(int i=1;i<=k;i++)
for(int j=1;j<=n;j++)
scanf("%d",&d[i].a[j]);
for(int i=0;i<=k;i++) build(i);
int ans=0;
for(int i=1;i<=n;i++){
bit t=d[0].get(i);
for(int j=0;j<=k;j++) t&=d[j].get(i);
ans+=t.count();
}
printf("%d
",ans);
return 0;
}
K-D Tree
距离最近的个点
点数量，
const int inf=1e4+5;
int key,root,n,m,q,k,mxd;
int sqr(int x)
{
return x*x;
}
struct point
{
int d[7];
friend int dis(point a,point b)
{
int s=0;
for (int i=0; i<k; i++)
s+=sqr(a.d[i]-b.d[i]);
return s;
}
} po;
struct node
{
point nw;
int son[2],mi[7],mx[7];
friend bool operator <(node a,node b)
{
return a.nw.d[key]<b.nw.d[key];
}
};
struct li
{
point a; int l;
friend bool operator <(li a, li b)
{
return a.l<b.l;
}
} mx;
set<li> st;
struct kdtree
{
node a[500010];
void init()
{
a[0].son[0]=a[0].son[1]=0;
for (int i=0; i<5; i++)
{
a[0].mx[i]=-inf;
a[0].mi[i]=inf;
}
}
void update(int x)
{
int l=a[x].son[0],r=a[x].son[1];
for (int i=0; i<k; i++)
{
a[x].mi[i]=min(a[x].nw.d[i],min(a[l].mi[i],a[r].mi[i]));
a[x].mx[i]=max(a[x].nw.d[i],max(a[l].mx[i],a[r].mx[i]));
}
}
int build(int l,int r,int cur)
{
if (l>r) return 0;
int m=(l+r)>>1;
key=cur; nth_element(a+l,a+m,a+r+1);
a[m].son[0]=build(l,m-1,(cur+1)%k);
a[m].son[1]=build(m+1,r,(cur+1)%k);
update(m);
return m;
}
int getmi(int x)
{
int s=0;
for (int i=0; i<k; i++)
s+=sqr(max(po.d[i]-a[x].mx[i],0)+max(a[x].mi[i]-po.d[i],0));
return s;
}
void ask(int q)
{
if (!q) return;
int tmp=dis(a[q].nw,po);
st.insert((li){a[q].nw,tmp});
if (st.size()>m)
{
set<li>::iterator it=st.end(); it--;
st.erase(it);
}
mxd=(*st.rbegin()).l;
int l=a[q].son[0],r=a[q].son[1],dl=2147483647,dr=2147483647;
if (l) dl=getmi(l);
if (r) dr=getmi(r);
if (dl<dr)
{
if (dl<mxd||st.size()<m) ask(l);
if (dr<mxd||st.size()<m) ask(r);
}
else {
if (dr<mxd||st.size()<m) ask(r);
if (dl<mxd||st.size()<m) ask(l);
}
}
} kd;
int main()
{
while (scanf("%d%d",&n,&k)!=EOF)
{
kd.init();
for (int i=1; i<=n; i++)
for (int j=0; j<k; j++)
scanf("%d",&kd.a[i].nw.d[j]);
root=kd.build(1,n,0);
scanf("%d",&q);
while (q--)
{
for (int i=0; i<k; i++)
scanf("%d",&po.d[i]);
scanf("%d",&m);
st.clear();
kd.ask(root);
printf("the closest %d points are:
",m);
for (set<li>::iterator it=st.begin(); it!=st.end(); it++)
{
point ans=(*it).a;
for (int i=0; i<k; i++)
{
printf("%d",ans.d[i]);
if (i!=k-1) printf(" "); else puts("");
}
}
}
}
}
k远点对（二维）
给出个点，求欧氏距离下的第远点对
#define N 100005
#define inf 0x7fffffff
int n,m,root,Q;
struct KDtree
{
int min[2],max[2],d[2],l,r;
ll dis;
bool operator < (const KDtree &a) const
{
return dis<a.dis;
}
}t[N];
priority_queue <KDtree> q;
void updata(int x,int y)
{
t[x].min[0]=min(t[x].min[0],t[y].min[0]);
t[x].max[0]=max(t[x].max[0],t[y].max[0]);
t[x].min[1]=min(t[x].min[1],t[y].min[1]);
t[x].max[1]=max(t[x].max[1],t[y].max[1]);
}
int build(int l,int r,int now)
{
for (int i=l;i<=r;i++)
t[i].dis=t[i].d[now];
int mid=(l+r)/2;
nth_element(t+l,t+mid,t+r+1);
t[mid].min[0]=t[mid].max[0]=t[mid].d[0];
t[mid].min[1]=t[mid].max[1]=t[mid].d[1];
if (l<mid) t[mid].l=build(l,mid-1,1-now);
if (r>mid) t[mid].r=build(mid+1,r,1-now);
if (t[mid].l) updata(mid,t[mid].l);
if (t[mid].r) updata(mid,t[mid].r);
return mid;
}
ll dist(int x,int y)
{
return (ll)(t[x].d[0]-t[y].d[0])*(t[x].d[0]-t[y].d[0])+(ll)(t[x].d[1]-t[y].d[1])*(t[x].d[1]-t[y].d[1]);
}
ll get(int x)
{
ll ret=0;
ret+=max((ll)(t[Q].d[0]-t[x].min[0])*(t[Q].d[0]-t[x].min[0]),(ll)(t[x].max[0]-t[Q].d[0])*(t[x].max[0]-t[Q].d[0]));
ret+=max((ll)(t[Q].d[1]-t[x].min[1])*(t[Q].d[1]-t[x].min[1]),(ll)(t[x].max[1]-t[Q].d[1])*(t[x].max[1]-t[Q].d[1]));
return ret;
}
void solve(int x)
{
t[x].dis=-dist(x,Q);
if (q.size()<m) q.push(t[x]);
else
{
KDtree u=q.top();
if (u.dis>t[x].dis)
{
q.pop();
q.push(t[x]);
}
}
ll dl=inf,dr=inf;
if (t[x].l) dl=-get(t[x].l);
if (t[x].r) dr=-get(t[x].r);
if (dl<dr)
{
if (dl<q.top().dis||q.size()<m&&t[x].l) solve(t[x].l);
if (dr<q.top().dis||q.size()<m&&t[x].r) solve(t[x].r);
}else
{
if (dr<q.top().dis||q.size()<m&&t[x].r) solve(t[x].r);
if (dl<q.top().dis||q.size()<m&&t[x].l) solve(t[x].l);
}
}
int main()
{
scanf("%d%d",&n,&m);
m*=2;
for (int i=1;i<=n;i++)
scanf("%d%d",&t[i].d[0],&t[i].d[1]);
root=build(1,n,0);
for (int i=1;i<=n;i++)
{
Q=i;
solve(root);
}
printf("%lld",-q.top().dis);
return 0;
}
维护二维区间和
1 x y A 在坐标增加权值A
2 x1 y1 x2 y2 输出x1 y1 x2 y2这个矩形内的数字和
3 终止程序运行
inline void read(int &x){
x=0;char ch;bool flag = false;
while(ch=getchar(),ch<'!');if(ch == '-') ch=getchar(),flag = true;
while(x=10*x+ch-'0',ch=getchar(),ch>'!');if(flag) x=-x;
}
inline int cat_max(const int &a,const int &b){return a>b ? a:b;}
inline int cat_min(const int &a,const int &b){return a<b ? a:b;}
const int maxn = 160010;
const int lim_siz = 2000;
int split[maxn],now;
int dem = 2;
struct Node{
int pos[2],val;
int minn[2],maxx[2],sum;
Node *ch[2];
void update(){
for(int d=0;d<dem;++d) minn[d] = min(pos[d],min(ch[0]->minn[d],ch[1]->minn[d]));
for(int d=0;d<dem;++d) maxx[d] = max(pos[d],max(ch[0]->maxx[d],ch[1]->maxx[d]));
sum = val + ch[0]->sum + ch[1]->sum;
}
}*null,*root,*op;
Node T[maxn];
inline bool cmp(const Node &a,const Node &b){
return a.pos[split[now]] < b.pos[split[now]];
}
inline void init(){
null = &T[0];
null->ch[0] = null->ch[1] = null;
null->val = 0;
for(int d=0;d<dem;++d){
null->pos[d] = 0;
null->minn[d] = 0x3f3f3f3f;
null->maxx[d] = -0x3f3f3f3f;
}root = null;
}
Node* build(int l,int r,int s){
if(l > r) return null;
int mid = (l+r)>> 1;
split[now = mid] = s % dem;
nth_element(T+l,T+mid,T+r+1,cmp);
Node *p = &T[mid];
p->ch[0] = build(l,mid-1,s+1);
p->ch[1] = build(mid+1,r,s+1);
p->update();return p;
}
int sav[maxn][2],sav_cnt,cnt;
int val[maxn],cmd,X1,Y1,X2,Y2,x;
int query(Node *p){
if(p == null) return 0;
if(p->minn[0] >= X1 && p->minn[1] >= Y1
&& p->maxx[0] <= X2 && p->maxx[1] <= Y2)
return p->sum;
else if(p->maxx[1] < Y1 || p->maxx[0] < X1
|| p->minn[0] > X2 || p->minn[1] > Y2)
return 0;
if( p->pos[0] >= X1 && p->pos[1] >= Y1
&& p->pos[0] <= X2 && p->pos[1] <= Y2)
return p->val + query(p->ch[0]) + query(p->ch[1]);
return query(p->ch[0]) + query(p->ch[1]);
}
int main(){
init();
int n;read(n);//read(n);
int lastans = 0;
while(1){
read(cmd);
if(cmd == 3) break;
if(cmd == 1){
read(X1);read(Y1);read(x);
X1 ^= lastans;Y1 ^= lastans;x ^= lastans;
sav[++sav_cnt][0] = X1;
sav[sav_cnt][1] = Y1;
val[sav_cnt] = x;
if(sav_cnt == lim_siz){
for(int i=1;i<=sav_cnt;++i){
T[cnt+i].minn[0] = T[cnt+i].maxx[0] = T[cnt+i].pos[0] = sav[i][0];
T[cnt+i].minn[1] = T[cnt+i].maxx[1] = T[cnt+i].pos[1] = sav[i][1];
T[cnt+i].val = val[i];
}root = build(1,cnt+=sav_cnt,1);
sav_cnt = 0;
}
}else{
read(X1);read(Y1);read(X2);read(Y2);
X1 ^= lastans;Y1 ^= lastans;
X2 ^= lastans;Y2 ^= lastans;
int ans = 0;
for(int i=1;i<=sav_cnt;++i){
if(sav[i][0] >= X1 && sav[i][1] >= Y1
&& sav[i][0] <= X2 && sav[i][1] <= Y2)
ans += val[i];
}
ans += query(root);
printf("%d
",lastans = ans);
}
}
return 0;
}
统计某一区间的最大值(三维)
在之间找一个在这个区间里只出现过一次且最大的数
对于每一个数维护三个维度：（自己的位置，相同数前一个位置，相同数后一个位置）
我们要的是且且且$ z>r$的最大数。
#define N 200003
int n,m,cmpd,root,pre[N],next[N],point[N],ans,lastans,x,y,val[N];
struct data {
int d[3],mx[3],mn[3],val,maxn;
int l,r;
}tr[N];
int tmp[10][10];
int cmp(data a,data b)
{
return a.d[cmpd]<b.d[cmpd];
}
void update(int now)
{
int l=tr[now].l; int r=tr[now].r;
for (int i=0;i<=2;i++){
if (l) tr[now].mx[i]=max(tr[now].mx[i],tr[l].mx[i]),
tr[now].mn[i]=min(tr[now].mn[i],tr[l].mn[i]);
if (r) tr[now].mx[i]=max(tr[now].mx[i],tr[r].mx[i]),
tr[now].mn[i]=min(tr[now].mn[i],tr[r].mn[i]);
}
if (l) tr[now].maxn=max(tr[l].maxn,tr[now].maxn);
if (r) tr[now].maxn=max(tr[r].maxn,tr[now].maxn);
}
int build(int l,int r,int d)
{
d%=3;
cmpd=d;
int mid=(l+r)/2;
nth_element(tr+l,tr+mid,tr+r+1,cmp);
for (int i=0;i<=2;i++)
tr[mid].mx[i]=tr[mid].mn[i]=tr[mid].d[i];
//cout<<tr[mid].d[0]<<" "<<tr[mid].d[1]<<" "<<tr[mid].d[2]<<endl;;
if (l<mid) tr[mid].l=build(l,mid-1,d+1);
if (r>mid) tr[mid].r=build(mid+1,r,d+1);
update(mid);
return mid;
}
int check(int now)
{
if (tr[now].mx[0]<x||tr[now].mn[0]>y) return 0;
if (tr[now].mn[1]>=x) return 0;
if (tr[now].mx[2]<=y) return 0;
return 1;
}
void query(int now)
{
if (tr[now].d[0]>=x&&tr[now].d[0]<=y&&tr[now].d[1]<x&&tr[now].d[2]>y)
ans=max(ans,tr[now].val);
int l=tr[now].l; int r=tr[now].r;
if (l&&tr[l].maxn>ans)
if (check(l)) query(l);
if (r&&tr[r].maxn>ans)
if (check(r)) query(r);
}
int main()
{
scanf("%d%d",&n,&m);
for (int i=1;i<=n;i++) scanf("%d",&val[i]);
memset(point,0,sizeof(point));
for (int i=1;i<=n;i++)
pre[i]=point[val[i]],point[val[i]]=i;
for (int i=1;i<=n;i++) point[i]=n+1;
for (int i=n;i>=1;i--)
next[i]=point[val[i]],point[val[i]]=i;
for (int i=1;i<=n;i++)
tr[i].d[0]=i,tr[i].d[1]=pre[i],tr[i].d[2]=next[i],tr[i].val=tr[i].maxn=val[i];
root=build(1,n,0);
lastans=0;
for (int i=1;i<=m;i++) {
scanf("%d%d",&x,&y);
x=(x+lastans)%n+1;
y=(y+lastans)%n+1;
if (x>y) swap(x,y);
ans=0;
query(root);
printf("%d
",ans);
lastans=ans;
}
}
倍增思想
一维倍增
int n,a[300010],f[300010][30],ans[300010];
int GCD(int x,int y)
{
int r=x%y;
while (r!=0) x=y,y=r,r=x%y;
return y;
}
void init()
{
for (int i=1;i<=n;i++)
f[i][0]=a[i];
for (int i=1;(1<<i)<=n;i++)
for (int j=1;j+(1<<i)-1<=n;j++)
f[j][i]=GCD(f[j][i-1],f[j+(1<<(i-1))][i-1]);
}
int query(int l,int r)
{
int k=(int)(log(double(r-l+1))/log((double)2));
return GCD(f[l][k],f[r-(1<<k)+1][k]);
}
int main()
{
cin>>n;
for (int i=1;i<=n;i++)
scanf("%d",&a[i]);
init();
return 0;
}
二维倍增
const int maxn=1010,lgn =11,INF =1e9;
int mymax(int a,int b,int c,int d){return max(max(max(a,b),c),d);}
int mymin(int a,int b,int c,int d){return min(min(min(a,b),c),d);}
int mx[maxn][maxn][lgn],mn[maxn][maxn][lgn];
int Log[maxn],A,B,n;
void st_pre()
{
for(int k=1;k<=Log[min(A,B)];k++)
for(int i=1;i+(1<<k)-1<=A;i++)
for(int j=1;j+(1<<k)-1<=B;j++)
{
mx[i][j][k]=mymax(mx[i][j][k-1],mx[i+(1<<k-1)][j][k-1],mx[i][j+(1<<k-1)][k-1],mx[i+(1<<k-1)][j+(1<<k-1)][k-1]);
mn[i][j][k]=mymin(mn[i][j][k-1],mn[i+(1<<k-1)][j][k-1],mn[i][j+(1<<k-1)][k-1],mn[i+(1<<k-1)][j+(1<<k-1)][k-1]);
}
}
int st_max(int x1,int y1,int x2,int y2)
{
int k=Log[n];
return mymax(mx[x1][y1][k],mx[x2-(1<<k)+1][y1][k],mx[x1][y2-(1<<k)+1][k],mx[x2-(1<<k)+1][y2-(1<<k)+1][k]);
}
int st_min(int x1,int y1,int x2,int y2)
{
int k=Log[n];
return mymin(mn[x1][y1][k],mn[x2-(1<<k)+1][y1][k],mn[x1][y2-(1<<k)+1][k],mn[x2-(1<<k)+1][y2-(1<<k)+1][k]);
}
int main()
{
scanf("%d%d%d",&A,&B,&n);
for(int i=1;i<=A;i++)
for(int j=1;j<=B;j++)
{
int x; scanf("%d",&x);
mx[i][j][0]=mn[i][j][0]=x;
}
Log[1]=0; for(int i=2;i<=max(A,B);i++) Log[i]=Log[i/2]+1;
st_pre();
int ans=INF;
for(int i=1;i+n-1<=A;i++)
for(int j=1;j+n-1<=B;j++)
ans=min(ans,st_max(i,j,i+n-1,j+n-1)-st_min(i,j,i+n-1,j+n-1));
printf("%d
",ans);
return 0;
}
树上倍增
int n,m,S,c[100010],h[100010],to[200010],ne[200010],d[100010],p[200010][20];
void dfs(int u)
{
c[u]=1;
for (int i=h[u];i!=-1;i=ne[i])
{
if (!d[to[i]])
{
d[to[i]]=d[u]+1;
p[to[i]][0]=u;
dfs(to[i]);
c[u]+=c[to[i]];
}
}
}
void init()
{
for (int j=1;(1<<j)<=n;j++)
for (int i=1;i<=n;i++)
if (p[i][j-1]!=-1) p[i][j]=p[p[i][j-1]][j-1];
}
int lca(int x,int y)
{
if (d[x]<d[y]) swap(x,y);
int i;
for (i=0;(1<<i)<=d[x];i++);i--;
for (int j=i;j>=0;j--)
if (d[x]-(1<<j)>=d[y]) x=p[x][j];
if (x==y) return x;
for (int j=i;j>=0;j--)
if (p[x][j]!=-1 && p[x][j]!=p[y][j])
x=p[x][j],y=p[y][j];
return p[x][0];
}
void Add(int x,int y)
{
S++;to[S]=x;ne[S]=h[y];h[y]=S;
S++;to[S]=y;ne[S]=h[x];h[x]=S;
}
int Find(int x,int y)
{
y=d[x]-y;
int i;
for (i=0;(1<<i)<=d[x];i++);i--;
for (int j=i;j>=0;j--)
if (d[x]-(1<<j)>=y) x=p[x][j];
return x;
}
int main()
{
cin>>n;m=n-1;
memset(h,-1,sizeof(h));
memset(d,0,sizeof(d));
S=0;
int x,y,q;
while(m--) scanf("%d%d",&x,&y),Add(x,y);
d[1]=1;dfs(1);init();
return 0;
}
Top Tree
const int N = 100010 * 2, inf = ~0U >> 1;
struct tag
{
int a, b; //ax+b
tag() { a = 1, b = 0; }
tag(int x, int y) { a = x, b = y; }
bool ex() { return a != 1 || b; }
tag operator+(const tag &x) { return tag(a * x.a, b * x.a + x.b); }
};
int atag(int x, tag y) { return x * y.a + y.b; }
struct data
{
int sum, minv, maxv, size;
data() { sum = size = 0, minv = inf, maxv =-inf; }
data(int x) { sum = minv = maxv = x, size = 1; }
data(int a, int b, int c, int d) { sum = a, minv = b, maxv = c, size = d; }
data operator+(const data &x)
{
return data(sum + x.sum, min(minv, x.minv), max(maxv, x.maxv), size + x.size);
}
};
data operator+(const data &a, const tag &b)
{
return a.size ? data(a.sum * b.a + a.size * b.b, atag(a.minv, b), atag(a.maxv, b), a.size) : a;
}
int f[N], son[N][4], a[N], tot, rt, rub, ru[N];
bool rev[N], in[N];
int val[N];
data csum[N], tsum[N], asum[N];
tag ctag[N], ttag[N];
bool isroot(int x, int t)
{
if (t)
return !f[x] || !in[f[x]] || !in[x];
return !f[x] || (son[f[x]][0] != x && son[f[x]][1] != x) || in[f[x]] || in[x];
}
void rev1(int x)
{
if (!x)
return;
swap(son[x][0], son[x][1]);
rev[x] ^= 1;
}
void tagchain(int x, tag p)
{
if (!x)
return;
csum[x] = csum[x] + p;
asum[x] = csum[x] + tsum[x];
val[x] = atag(val[x], p);
ctag[x] = ctag[x] + p;
}
void tagtree(int x, tag p, bool t)
{
if (!x)
return;
tsum[x] = tsum[x] + p;
ttag[x] = ttag[x] + p;
if (!in[x] && t)
tagchain(x, p);
else
asum[x] = csum[x] + tsum[x];
}
void pb(int x)
{
if (!x)
return;
if (rev[x])
rev1(son[x][0]), rev1(son[x][1]), rev[x] = 0;
if (!in[x] && ctag[x].ex())
{
tagchain(son[x][0], ctag[x]);
tagchain(son[x][1], ctag[x]);
ctag[x] = tag();
}
if (ttag[x].ex())
{
tagtree(son[x][0], ttag[x], 0), tagtree(son[x][1], ttag[x], 0);
tagtree(son[x][2], ttag[x], 1), tagtree(son[x][3], ttag[x], 1);
ttag[x] = tag();
}
}
void up(int x)
{
tsum[x] = data();
for (int i = 0; i < 2; i++)
if (son[x][i])
tsum[x] = tsum[x] + tsum[son[x][i]];
for (int i = 2; i < 4; i++)
if (son[x][i])
tsum[x] = tsum[x] + asum[son[x][i]];
if (in[x])
{
csum[x] = data();
asum[x] = tsum[x];
}
else
{
csum[x] = data(val[x]);
for (int i = 0; i < 2; i++)
if (son[x][i])
csum[x] = csum[x] + csum[son[x][i]];
asum[x] = csum[x] + tsum[x];
}
}
int child(int x, int t)
{
pb(son[x][t]);
return son[x][t];
}
void rotate(int x, int t)
{
int y = f[x], w = (son[y][t + 1] == x) + t;
son[y][w] = son[x][w ^ 1];
if (son[x][w ^ 1])
f[son[x][w ^ 1]] = y;
if (f[y])
for (int z = f[y], i = 0; i < 4; i++)
if (son[z][i] == y)
son[z][i] = x;
f[x] = f[y];
f[y] = x;
son[x][w ^ 1] = y;
up(y);
}
void splay(int x, int t = 0)
{
int s = 1, i = x, y;
a[1] = i;
while (!isroot(i, t))
a[++s] = i = f[i];
while (s)
pb(a[s--]);
while (!isroot(x, t))
{
y = f[x];
if (!isroot(y, t))
{
if ((son[f[y]][t] == y) ^ (son[y][t] == x))
rotate(x, t);
else
rotate(y, t);
}
rotate(x, t);
}
up(x);
}
int newnode()
{
int x = rub ? ru[rub--] : ++tot;
son[x][2] = son[x][3] = 0;
in[x] = 1;
return x;
}
void setson(int x, int t, int y)
{
son[x][t] = y;
f[y] = x;
}
int pos(int x)
{
for (int i = 0; i < 4; i++)
if (son[f[x]][i] == x)
return i;
return 4;
}
void add(int x, int y)
{
if (!y)
return;
pb(x);
for (int i = 2; i < 4; i++)
if (!son[x][i])
{
setson(x, i, y);
return;
}
while (son[x][2] && in[son[x][2]])
x = child(x, 2);
int z = newnode();
setson(z, 2, son[x][2]);
setson(z, 3, y);
setson(x, 2, z);
splay(z, 2);
}
void del(int x)
{
if (!x)
return;
splay(x);
if (!f[x])
return;
int y = f[x];
if (in[y])
{
int s = 1, i = y, z = f[y];
a[1] = i;
while (!isroot(i, 2))
a[++s] = i = f[i];
while (s)
pb(a[s--]);
if (z)
{
setson(z, pos(y), child(y, pos(x) ^ 1));
splay(z, 2);
}
ru[++rub] = y;
}
else
{
son[y][pos(x)] = 0;
splay(y);
}
f[x] = 0;
}
int fa(int x)
{
splay(x);
if (!f[x])
return 0;
if (!in[f[x]])
return f[x];
int t = f[x];
splay(t, 2);
return f[t];
}
int access(int x)
{
int y = 0;
for (; x; y = x, x = fa(x))
{
splay(x);
del(y);
add(x, son[x][1]);
setson(x, 1, y);
up(x);
}
return y;
}
int lca(int x, int y)
{
access(x);
return access(y);
}
int root(int x)
{
access(x);
splay(x);
while (son[x][0])
x = son[x][0];
return x;
}
void makeroot(int x)
{
access(x);
splay(x);
rev1(x);
}
void link(int x, int y)
{
makeroot(x);
add(y, x);
access(x);
}
void cut(int x)
{
access(x);
splay(x);
f[son[x][0]] = 0;
son[x][0] = 0;
up(x);
}
void changechain(int x, int y, tag p)
{
makeroot(x);
access(y);
splay(y);
tagchain(y, p);
}
data askchain(int x, int y)
{
makeroot(x);
access(y);
splay(y);
return csum[y];
}
void changetree(int x, tag p)
{
access(x);
splay(x);
val[x] = atag(val[x], p);
for (int i = 2; i < 4; i++)
if (son[x][i])
tagtree(son[x][i], p, 1);
up(x);
splay(x);
}
data asktree(int x)
{
access(x);
splay(x);
data t = data(val[x]);
for (int i = 2; i < 4; i++)
if (son[x][i])
t = t + asum[son[x][i]];
return t;
}
int n, m, x, y, z, k, i, ed[N][2];
int main()
{
read(n); //n个点
read(m); //m个询问
tot = n;
for (i = 1; i < n; i++) //边
read(ed[i][0]), read(ed[i][1]);
for (i = 1; i <= n; i++) //点权
read(val[i]), up(i);
for (i = 1; i < n; i++)
link(ed[i][0], ed[i][1]);
read(rt);
makeroot(rt);
while (m--)
{
read(k);
if (k == 1)
{//换根
read(rt);
makeroot(rt);
}
if (k == 9)
{ //x的父亲变成y
read(x), read(y);
if (lca(x, y) == x) continue;
cut(x);
link(y, x);
makeroot(rt);
}
if (k == 0)
{ //子树赋值
read(x), read(y);
changetree(x, tag(0, y));
}
if (k == 5)
{ //子树加
read(x), read(y);
changetree(x, tag(1, y));
}
if (k == 3)
{ //子树最小值
read(x);
printf("%d
", asktree(x).minv);
}
if (k == 4)
{ //子树最大值
read(x);
printf("%d
", asktree(x).maxv);
}
if (k == 11)
{ //子树和
read(x);
printf("%d
", asktree(x).sum);
}
if (k == 2)
{ //链赋值
read(x), read(y), read(z);
changechain(x, y, tag(0, z));
makeroot(rt);
}
if (k == 6)
{ //链加
read(x), read(y), read(z);
changechain(x, y, tag(1, z));
makeroot(rt);
}
if (k == 7)
{ //链最小值
read(x), read(y);
printf("%d
", askchain(x, y).minv);
makeroot(rt);
}
if (k == 8)
{ //链最大值
read(x), read(y);
printf("%d
", askchain(x, y).maxv);
makeroot(rt);
}
if (k == 10)
{ //链和
read(x), read(y);
printf("%d
", askchain(x, y).sum);
makeroot(rt);
}
}
}
字符串处理
AC自动机
#include<cstring>
#include<queue>
#include<cstdio>
#include<map>
#include<string>
using namespace std;
const int SIGMA_SIZE = 26;
const int MAXNODE = 11000;
const int MAXS = 150 + 10;
struct AhoCorasickAutomata {
int ch[MAXNODE][SIGMA_SIZE];
int f[MAXNODE]; // fail函数
int val[MAXNODE]; // 每个字符串的结尾结点都有一个非0的val
int last[MAXNODE]; // 输出链表的下一个结点
int match[MAXNODE]; // 表示这个点是结点
int cnt[MAXS]; //用来统计模式串被找到了几次
int sz;
void init() {
sz = 1;
memset(ch[0], 0, sizeof(ch[0]));
memset(cnt, 0, sizeof(cnt));
memset(match, 0, sizeof(match));
}
// 字符c的编号
int idx(char c) {
return c-'a';
}
// 插入字符串。v必须非0
void insert(char *s, int v) {
int u = 0, n = strlen(s);
for(int i = 0; i < n; i++) {
int c = idx(s[i]);
if(!ch[u][c]) {
memset(ch[sz], 0, sizeof(ch[sz]));
val[sz] = 0;
ch[u][c] = sz++;
}
u = ch[u][c];
}
val[u] = v;
}
// 递归打印以结点j结尾的所有字符串
void print(int j) {
if(j) {
cnt[val[j]]++;
//match[j] = 1;
print(last[j]);
}
}
// 在T中找模板
int find(char* T) {
int n = strlen(T);
int j = 0; // 当前结点编号，初始为根结点
for(int i = 0; i < n; i++) { // 文本串当前指针
int c = idx(T[i]);
j = ch[j][c];
if(val[j]) print(j);
else if(last[j]) print(last[j]); // 找到了！
}
}
// 计算fail函数
void getFail() {
queue<int> q;
f[0] = 0;
// 初始化队列
for(int c = 0; c < SIGMA_SIZE; c++) {
int u = ch[0][c];
if(u) { f[u] = 0; q.push(u); last[u] = 0; }
}
// 按BFS顺序计算fail
while(!q.empty()) {
int r = q.front(); q.pop();
for(int c = 0; c < SIGMA_SIZE; c++) {
int u = ch[r][c];
if(!u) {ch[r][c] = ch[f[r]][c];continue;}
q.push(u);
int v = f[r];
while(v && !ch[v][c]) v = f[v];
f[u] = ch[v][c];
last[u] = val[f[u]] ? f[u] : last[f[u]];
}
}
/* *when Matrix need
for(int i = 0; i < sz; i++) {
if(val[i]) print(i);
else if(last[i]) print(i);
}
*/
/* 统计长度为n的串有多种可能不出现模板串,需要Matrix
int doit(int n) {
matrix A(sz, sz);
for(int i = 0; i < sz; i++) {
if(match[i]) continue;
for(int c = 0; c < SIGMA_SIZE; c++) {
if(!match[ch[i][c]]) A[i][ch[i][c]]++;
}
}
A = A ^ n;
int ans = 0;
for(int i = 0; i < sz; i++) {
ans += A[0][i];
ans %= MOD;
}
return ans;
}
*/
}
};
AhoCorasickAutomata ac;
char text[1000001], P[151][80];
int n, T;
map<string,int> ms;
int main()
{
while(scanf("%d", &n) == 1 && n)
{
ac.init();
for(int i = 1; i <= n; i++)
{
scanf("%s", P[i]);
ac.insert(P[i], i);
}
ac.getFail();
scanf("%s", text);
ac.find(text);
int best = -1;
for(int i = 1; i <= n; i++)
if(ac.cnt[i] > best) best = ac.cnt[i];
printf("%d
", best);
for(int i = 1; i <= n; i++)
if(ac.cnt[ms[string(P[i])]] == best) printf("%s
", P[i]);
}
return 0;
}
KMP
char a1[2000000],a2[2000000];
int kmp[2000000];
int main()
{
scanf("%s%s",a1,a2);
kmp[0]=kmp[1]=0;//前一位，两位失配了，都只可能将第一位作为新的开头
int len1=strlen(a1),len2=strlen(a2);
int k;
k=0;
for(int i=1;i<len2;i++)//自己匹配自己
{
while(k&&a2[i]!=a2[k])k=kmp[k];//找到最长的前后缀重叠长度
kmp[i+1]=a2[i]==a2[k]?++k:0;//不相等的情况，即无前缀能与后缀重叠，直接赋值位0（注意是给下一位，因为匹配的是下一位适失配的情况）
}
k=0;
for(int i=0;i<len1;i++)
{
while(k&&a1[i]!=a2[k])k=kmp[k];//如果不匹配，则将利用kmp数组往回跳
k+=a1[i]==a2[k]?1:0;//如果相等了，则匹配下一位
if(k==len2)printf("%d
",i-len2+2);//如果已经全部匹配完毕，则输出初始位置
}
for(int i=1;i<=len2;i++)printf("%d ",kmp[i]);//输出f数组
return 0;
}
拓展KMP
求s1的前缀和s2的后缀最大匹配。
#define maxn 111111
char a[maxn],b[maxn];
int nex[maxn];
int main()
{
while(scanf("%s%s",a,b)!=EOF)
{
int len1=strlen(a);
int len2=strlen(b);
int len=min(len1,len2);//最大匹配长度必须不大于两串串长较小值
for(int i=len1,j=0;i<len1+len2;i++,j++)//将a串b串合并
a[i]=b[j];
int la=len1+len2;//总串长
for(int i=0,j=-1;i<=la;i++,j++)//求next数组
{
nex[i]=j;
while(~j&&a[i]!=a[j])
j=nex[j];
}
while(nex[la]>len)
la=nex[la];
if(nex[la]==0)//最大匹配长度为0直接输出0
printf("0
");
else
{
for(int i=0;i<nex[la];i++)//输出最大匹配
printf("%c",a[i]);
printf(" %d
",nex[la]);//输出最大匹配长度
}
}
return 0;
}
后缀数组
第一行个整数，第个整数表示排名为的后缀的第一个字符在原串中的位置。
第二行个整数，第个整数表示排名为和排名为的后缀的最长公共前缀的长度。
时间复杂度
int n,m,s,b[200010],c[200010],d[200010];
char sx[200010],sy[200010];
struct data
{
int len,fa,letter[26],tree[26],id,flag;
}a[200010];
int sa[200010],rank[200010],RANK;
void Extend(int x,int p)
{
s++;int q=s;a[q].len=a[p].len+1;
while (p!=0 && a[p].letter[x]==0)
a[p].letter[x]=q,p=a[p].fa;
if (p==0) {a[q].fa=1;return ;}
int np=a[p].letter[x];
if (a[np].len==a[p].len+1) a[q].fa=np;
else
{
s++;int nq=s;a[nq].len=a[p].len+1;
for (int i=0;i<26;i++)
a[nq].letter[i]=a[np].letter[i];
a[nq].id=a[np].id;
a[nq].fa=a[np].fa;a[np].fa=nq;a[q].fa=nq;
while (p!=0&&a[p].letter[x]==np)
a[p].letter[x]=nq,p=a[p].fa;
}
}
void Insert(char x[])
{
int y=strlen(x);
s=1;int z=1;
for (int i=y-1;i>=0;i--)
Extend(x[i]-'a',z),z=a[z].letter[x[i]-'a'],a[z].id=i+1,d[i+1]=z,a[z].flag=1;
}
void dfs(int x)
{
if(a[x].id!=0 && a[x].flag) sa[++RANK]=a[x].id,rank[a[x].id]=RANK;
for (int i=0;i<26;i++)
if (a[x].tree[i]!=0) dfs(a[x].tree[i]);
}
void build()
{
for (int i=1;i<=s;i++)
c[a[i].len]++;
for (int i=1;i<=s;i++)
c[i]+=c[i-1];
for (int i=1;i<=s;i++)
b[c[a[i].len]--]=i;
for (int i=s;i>=1;i--)
{
int p=b[i];
a[a[p].fa].tree[sx[a[p].id+a[a[p].fa].len-1]-'a']=p;
}
RANK=0;
dfs(1);
}
LL Query(int x,int y)
{
if (x==y) return a[x].len;
if (a[x].len>a[y].len) return Query(a[x].fa,y);
else return Query(x,a[y].fa);
}
int main()
{
n=read(sx);n=strlen(sx);
Insert(sx);
build();
for (int i=1;i<n;i++)
print(sa[i]),putchar(' ');
print(sa[n]);puts("");
for (int i=1;i<n-1;i++)
print(Query(d[sa[i]],d[sa[i+1]])),putchar(' ');
if (n>1) print(Query(d[sa[n-1]],d[sa[n]])),puts("");
return 0;
}
后缀自动机
最长公共子串（多串）
int n,m,s,b[500010],c[500010];
char sx[250010],sy[250010];
struct data
{
int len,fa,letter[26],res,x;
}a[500010];
const int inf=500010;
void Extend(int x,int p)
{
s++;int q=s;a[q].len=a[p].len+1;
while (p!=0 && a[p].letter[x]==0)
a[p].letter[x]=q,p=a[p].fa;
if (p==0) {a[q].fa=1;return ;}
int np=a[p].letter[x];
if (a[np].len==a[p].len+1) a[q].fa=np;
else
{
s++;int nq=s;a[nq].len=a[p].len+1;
for (int i=0;i<26;i++)
a[nq].letter[i]=a[np].letter[i];
a[nq].fa=a[np].fa;a[np].fa=nq;a[q].fa=nq;
while (p!=0&&a[p].letter[x]==np)
a[p].letter[x]=nq,p=a[p].fa;
}
}
void Insert(char x[])
{
int y=strlen(x);
s=1;int z=1;
for (int i=0;i<y;i++)
Extend(x[i]-'a',z),z=a[z].letter[x[i]-'a'];
memset(c,0,sizeof(c));
for (int i=1;i<=s;i++)
a[i].res=a[i].len,c[a[i].len]++;
for (int i=1;i<=s;i++)
c[i]+=c[i-1];
for (int i=1;i<=s;i++)
b[c[a[i].len]--]=i;
}
void Query(char x[])
{
int res=0,p=1,z,y=strlen(x);
for (int i=0;i<y;i++)
{
z=x[i]-'a';
while (p!=0&&a[p].letter[z]==0) p=a[p].fa,res=a[p].len;
if (a[p].letter[z]!=0) res++,p=a[p].letter[z];
else res=0,p=1;
a[p].x=max(a[p].x,res);
}
for (int i=s;i>=1;i--)
{
p=b[i];
a[a[p].fa].x=max(a[a[p].fa].x,a[p].x);
a[p].res=min(a[p].res,a[p].x);
a[p].x=0;
}
}
int main()
{
n=read(sx);
Insert(sx);
while (read(sy))
Query(sy);
int ans=0;
for (int i=1;i<=s;i++)
ans=max(ans,a[i].res);
print(ans),puts("");
return 0;
}
求第小字符串
不同位置相同子串算多个
int n,m,s,b[1000010],c[1000010];
char sx[1000010],sy[1000010];
struct data
{
int fa,letter[26];
LL len,child,res;
}a[1000010];
void Extend(int x,int p)
{
s++;int q=s;a[q].len=a[p].len+1;
while (p!=0 && a[p].letter[x]==0)
a[p].letter[x]=q,p=a[p].fa;
if (p==0) {a[q].fa=1;return ;}
int np=a[p].letter[x];
if (a[np].len==a[p].len+1) a[q].fa=np;
else
{
s++;int nq=s;a[nq].len=a[p].len+1;
for (int i=0;i<26;i++)
a[nq].letter[i]=a[np].letter[i];
a[nq].fa=a[np].fa;a[np].fa=nq;a[q].fa=nq;
while (p!=0&&a[p].letter[x]==np)
a[p].letter[x]=nq,p=a[p].fa;
}
}
void Insert(char x[])
{
int y=strlen(x);
s=1;int z=1;
for (int i=0;i<y;i++)
Extend(x[i]-'a',z),z=a[z].letter[x[i]-'a'],a[z].child++;
memset(c,0,sizeof(c));
for (int i=1;i<=s;i++)
a[i].res=0,c[a[i].len]++;
for (int i=1;i<=s;i++)
c[i]+=c[i-1];
for (int i=1;i<=s;i++)
b[c[a[i].len]--]=i;
}
void init()
{
for (int i=s;i>=1;i--)
{
int p=b[i];
if (n==1) a[a[p].fa].child+=a[p].child;
else a[p].child=1;
}
for (int i=s;i>=1;i--)
{
int p=b[i];a[p].res=a[p].child;
for (int j=0;j<26;j++)
a[p].res+=a[a[p].letter[j]].res;
}
}
void Query(int y,int x)
{
if (x<=0) return ;
int i=0;
while(i<26&&a[a[y].letter[i]].res<x)
x-=a[a[y].letter[i]].res,i++;
if (i==26) {puts("-1");return ;}
else {putchar(i+'a'),Query(a[y].letter[i],x-a[a[y].letter[i]].child);}
}
int main()
{
n=read(sx);
Insert(sx);
read(n);read(m);
init();
a[1].child=0;
Query(1,m);
return 0;
}
不同位置相同子串算一个
namespace SuffixAutomation {
const int MAXN = 180005;
struct Node {
Node *next[26], *fa;
int max, cnt;
Node(int max = 0) : max(max) {}
Node(int max, Node *p) {
*this = *p, this->max = max;
}
inline void *operator new(size_t);
} pool[MAXN], *cur = pool, *root, *last;
inline void *Node::operator new(size_t) {
return cur++;
}
inline void init() {
root = last = new Node();
}
inline Node *extend(int c, Node *p) {
Node *np = new Node(p->max + 1);
while (p && !p->next[c]) p->next[c] = np, p = p->fa;
if (!p) {
np->fa = root;
} else {
Node *q = p->next[c];
if (q->max == p->max + 1) {
np->fa = q;
} else {
Node *nq = new Node(p->max + 1, q);
q->fa = np->fa = nq;
while (p && p->next[c] == q) p->next[c] = nq, p = p->fa;
}
}
return np;
}
std::vector<Node *> buc[MAXN];
typedef std::pair<Node *, int> Pair;
std::vector<Pair> edge[MAXN];
/*预处理从每个节点出发，还有多少不同的子串可以到达。*/
inline void prepare(const char *s, const int n) {
for (Node *p = pool; p != cur; p++) p->cnt = 1, buc[p->max].push_back(p);
for (register int i = n; i >= 0; i--) {
for (auto p : buc[i]) {
for (register int j = 0; j < 26; j++) {
if (p->next[j]) {
edge[p - pool].push_back(Pair(p->next[j], j));
/*这里是预处理，保证时间复杂度，并不是 endPos 集合，记录的是当前节点不同子串个数*/
p->cnt += p->next[j]->cnt;
}
}
}
}
}
inline void dfs(Node *p, int k) {
if (--k == 0) return;
for (auto i : edge[p - pool]) {
if (i.first->cnt < k) {
k -= i.first->cnt;
} else {
putchar('a' + i.second), dfs(i.first, k);
break;
}
}
}
inline void solve() {
static char s[MAXN];
scanf("%s", s);
register int n = strlen(s);
init();
for(register int i = 0; i < n; i++) last = extend(s[i] - 'a', last);
prepare(s, n);
register int q;
scanf("%d", &q);
for (register int i = 1, k; i <= q; i++) {
scanf("%d", &k), dfs(root, k + 1), putchar('
');
}
}
}
int main() {
SuffixAutomation::solve();
}
不同子串的出现次数
指长度为的串出现次数的最大值，输出…
一个节点的出现次数是，一个节点的长度范围是
namespace SuffixAutomation {
const int MAXN = 500005;
struct Node {
Node *next[26], *fa;
int max, endPosSize;
Node(int max = 0) : max(max) {}
Node(int max, Node *p) {
*this = *p, this->max = max;
}
inline void *operator new(size_t);
} pool[MAXN], *cur = pool, *root, *last;
inline void *Node::operator new(size_t) {
return cur++;
}
inline void init() {
root = last = new Node();
}
inline Node *extend(int c, Node *p) {
Node *np = new Node(p->max + 1);
while (p && !p->next[c]) p->next[c] = np, p = p->fa;
if (!p) {
np->fa = root;
} else {
Node *q = p->next[c];
if (q->max == p->max + 1) {
np->fa = q;
} else {
Node *nq = new Node(p->max + 1, q);
q->fa = np->fa = nq;
while (p && p->next[c] == q) p->next[c] = nq, p = p->fa;
}
}
return np;
}
inline void getEndPosSize(const char *s, const int n) {
/*将Suffix Link 上的所有叶子节点的endPosSize设置为1*/
Node *p = root;
for (register int i = 0; i < n; i++)
p = p->next[*s++ - 'a'], p->endPosSize++;
/*按照 len 从大到小的顺序更新*/
static std::vector<Node *> buc[MAXN];
static int ans[MAXN];
for (Node *p = pool; p != cur; p++) buc[p->max].push_back(p);
for (register int i = n; i; i--) {
for (auto p : buc[i]) {
ans[i] = std::max(ans[i], p->endPosSize); //这里做更新
p->fa->endPosSize += p->endPosSize;
}
}
for (register int i = n - 1; i; i--) ans[i] = std::max(ans[i], ans[i + 1]); //这道题目的特殊优化
for (register int i = 1; i <= n; i++) std::cout << ans[i] << "
";
}
inline void solve() {
init();
static char s[MAXN];
scanf("%s", s);
register int n = strlen(s);
for (register int i = 0; i < n; i++)
last = extend(s[i] - 'a', last);
getEndPosSize(s, n);
}
}
int main() {
SuffixAutomation::solve();
return 0;
}
最长不重叠重复子串
“主题”是整个音符序列的一个子串，它需要满足如下条件：
1.长度至少为 5 个音符。
2.在乐曲中重复出现。(可能经过转调，“转调”的意思是主题序列中每个音符都被加上或减去了同一个整数值)
3.重复出现的同一主题不能有公共部分。
对于转调，做一次差分，然后回归此问题
namespace SuffixAutomation {
const int MAXN = 20005;
struct Node {
Node *fa, *next[175];
int max, minPos, maxPos;
Node(int max = 0) : max(max), minPos(20000), maxPos(0), fa(NULL) {
memset(next, 0, sizeof(next));
}
Node(int max, Node *p) {
*this = *p, this->max = max;
}
inline void *operator new(size_t);
} pool[MAXN << 1], *cur = pool, *root, *last;
int n, a[MAXN];
std::vector<Node *> buc[MAXN];
inline void *Node::operator new(size_t) {
return cur++;
}
inline void init() {
cur = pool;
root = last = new Node();
for (register int i = 0; i <= n; i++) buc[i].clear();
}
inline Node *extend(int c, Node *p) {
Node *np = new Node(p->max + 1);
while (p && !p->next[c]) p->next[c] = np, p = p->fa;
if (!p) {
np->fa = root;
} else {
Node *q = p->next[c];
if (q->max == p->max + 1) {
np->fa = q;
} else {
Node *nq = new Node(p->max + 1, q);
q->fa = np->fa = nq;
while (p && p->next[c] == q) p->next[c] = nq, p = p->fa;
}
}
return np;
}
inline int getAns() {
register int res = 0;
Node *p = root;
p->minPos = p->maxPos = 0;
for (register int i = 0; i < n; i++)
p = p->next[a[i]], p->maxPos = p->minPos = i + 1;
for (Node *i = pool; i != cur; i++) buc[i->max].push_back(i);
for (register int i = n; i; i--) {
for (register int j = 0; j < buc[i].size(); j++) {
Node *tmp = buc[i][j];
res = std::max(res, std::min(tmp->max, tmp->maxPos - tmp->minPos - 1));
tmp->fa->minPos = std::min(tmp->fa->minPos, tmp->minPos);
tmp->fa->maxPos = std::max(tmp->fa->maxPos, tmp->maxPos);
}
}
return res < 4 ? 0 : res + 1;
}
inline void solve() {
while (scanf("%d", &n), n) {
for (register int i = 0; i < n; i++) scanf("%d", a + i);
n--;
for (register int i = 0; i < n; i++) a[i] = a[i + 1] - a[i] + 87;
init();
for (register int i = 0; i < n; i++) last = extend(a[i], last);
std::cout << getAns() << "
";
}
}
}
int main() {
SuffixAutomation::solve();
return 0;
}
本质不同的子串个数
namespace SuffixAutomation {
const int MAXN = 50000;
struct Node {
Node *fa, *next[128];
int max;
Node(int max = 0) : max(max), fa(NULL) {
memset(next, 0, sizeof(next));
}
Node(int max, Node *q) {
*this = *q, this->max = max;
}
inline void *operator new(size_t);
} pool[MAXN + 1 << 1], *cur = pool, *root, *last;
inline void *Node::operator new(size_t) {
return cur++;
}
int ans;
inline void init() {
cur = pool;
ans = 0;
root = last = new Node();
}
inline Node *extend(int c, Node *p) {
Node *np = new Node(p->max + 1);
while (p && !p->next[c]) p->next[c] = np, p = p->fa;
if (!p) {
np->fa = root;
} else {
Node *q = p->next[c];
if (q->max == p->max + 1) {
np->fa = q;
} else {
Node *nq = new Node(p->max + 1, q);
q->fa = np->fa = nq;
while (p && p->next[c] == q) p->next[c] = nq, p = p->fa;
}
}
ans += np->max - np->fa->max;
return np;
}
inline void solve() {
register int n;
scanf("%d", &n);
for (register int i = 0; i < n; i++) {
static char s[MAXN];
scanf("%s", s);
init();
register int len = strlen(s);
for (register int i = 0; i < len; i++) last = extend(s[i], last);
std::cout << ans << "
";
}
}
}
int main() {
SuffixAutomation::solve();
return 0;
}
后缀平衡树
一棵根节点为的树，每个节点代表一个字符
对于每个节点求这个节点到根节点路径上的不同子串的个数
输出出N个数，表示节点i到根节点路径上的不同子串的个数
const int N=1e5+10;
const int bas=31;
int hs[N],M[N];
int n,len,ans,Ans[N];
inline int read(){
int f=1,x=0;char ch;
do{ch=getchar();if(ch=='-')f=-1;}while(ch<'0'||ch>'9');
do{x=x*10+ch-'0';ch=getchar();}while(ch>='0'&&ch<='9');
return f*x;
}
struct Suffix_Balanced_ScapeGoat_Tree{
int lx[N],rx[N],rt,size[N],s[N],cnt,Q[N],tail;
ll tag[N];
int get(int i,int l){return hs[i]-hs[i-l]*M[l];}
int qlcp(int x,int y){
int l=0,r=min(x,y);
while(l<r){
int mid=(l+r)>>1;
if(get(x,mid+1)==get(y,mid+1))l=mid+1;else r=mid;
}
return l;
}
inline bool cmp(int x,int y){int l=qlcp(x,y);return s[x-l]<s[y-l];}
inline int merge(int x,int y){
if(!x||!y)return x|y;
if(size[x]>size[y]){size[x]+=size[y];rx[x]=merge(rx[x],y);return x;}
else{size[y]+=size[x];lx[y]=merge(x,lx[y]);return y;}
}
inline int build(int ls,int rs,ll l,ll r){
if(ls>rs)return 0;
int mid=(ls+rs)>>1;ll midv=(l+r)>>1;int x=Q[mid];tag[x]=midv;
lx[x]=build(ls,mid-1,l,midv);rx[x]=build(mid+1,rs,midv,r);
size[x]=rs-ls+1;return x;
}
inline void dfs(int x){if(lx[x])dfs(lx[x]);Q[++tail]=x;if(rx[x])dfs(rx[x]);}
inline int rebuild(int x,ll l,ll r){
tail=0;dfs(x);return build(1,tail,l,r);
}
inline int ins(int x,ll l,ll r,int val){
if(!x){
size[++cnt]=1;lx[cnt]=rx[cnt]=0;tag[cnt]=(l+r)>>1;
return cnt;
}
size[x]++;
if(cmp(x,val)){
rx[x]=ins(rx[x],tag[x],r,val);
if(size[rx[x]]>0.65*size[x])x=rebuild(x,l,r);
}
else{
lx[x]=ins(lx[x],l,tag[x],val);
if(size[lx[x]]>0.65*size[x])x=rebuild(x,l,r);
}
return x;
}
inline int del(int x,int val){
if(x==val)return merge(lx[x],rx[x]);
size[x]--;
if(tag[x]<tag[val])rx[x]=del(rx[x],val);
else lx[x]=del(lx[x],val);
return x;
}
inline int queryrk(int key){
int x=rt,ans=0;
while(1){
int i=size[lx[x]]+1;
if(key==x)return ans+i;
if(tag[x]<tag[key])x=rx[x],ans+=i;else x=lx[x];
}
}
inline int find(int key){
int x=rt;
while(1){
int i=size[lx[x]]+1;
if(key==i)return x;
if(key>i)x=rx[x],key-=i;else x=lx[x];
}
}
inline void del(int x){
int rk=queryrk(x),y=find(rk-1),z=find(rk+1);
ans-=qlcp(x,y)+qlcp(x,z)-qlcp(y,z);
rt=del(rt,x);cnt--;len--;
}
inline void ins(int x){
s[++len]=x;hs[len]=hs[len-1]*bas+x;
rt=ins(rt,0,1LL<<62,len);
if(len<3)return;
int rk=queryrk(len),y=find(rk-1),z=find(rk+1);
ans+=qlcp(len,y)+qlcp(len,z)-qlcp(y,z);
}
}T;
char str[N];
struct Edge{int u,v,next;}G[N<<1];int head[N],tot=0;
inline void addedge(int u,int v){
G[++tot].u=u;G[tot].v=v;G[tot].next=head[u];head[u]=tot;
G[++tot].u=v;G[tot].v=u;G[tot].next=head[v];head[v]=tot;
}
inline void dfs(int u,int fa){
T.ins(str[u]-'a'+1);Ans[u]=(len-1)*(len-2)/2-ans;
for(int i=head[u];i;i=G[i].next){
if(G[i].v!=fa)dfs(G[i].v,u);
}
T.del(len);
}
int main(){
freopen("balsuffix.in","r",stdin);
freopen("balsuffix.out","w",stdout);
M[0]=1;
for(int i=1;i<N;i++)M[i]=M[i-1]*bas;
T.ins(27);T.ins(0);int T=read();
while(T--){
n=read();for(int i=1;i<=n;i++)head[i]=0;tot=0;
for(int i=1;i<n;i++){
int u=read(),v=read();addedge(u,v);
}
scanf("%s",str+1);
dfs(1,0);
for(int i=1;i<=n;i++)printf("%d
",Ans[i]);
}
}
Palindromic树
const int maxn = 300010*2;
const int ALP = 26;
struct PAM{
int next[maxn][ALP];
int fail[maxn] ;//fail指针，失配后跳转到fail指针指向的节点
int cnt[maxn] ; //表示节点i表示的本质不同的串的个数（建树时求出的不是完全的，最后count()函数跑一遍以后才是正确的）
int num[maxn] ; //表示以节点i表示的最长回文串的最右端点为回文串结尾的回文串个数
int len[maxn] ;//len[i]表示节点i表示的回文串的长度（一个节点表示一个回文串）
int s[maxn] ;//存放添加的字符
int last ;//指向新添加一个字母后所形成的最长回文串表示的节点。
int n ;//表示添加的字符个数。
int p ;//表示添加的节点个数。
int newnode(int l){
for(int i=0;i<ALP;i++)
next[p][i]=0;
cnt[p]=num[p]=0;
len[p]=l;
return p++;
}
void init(){
p = 0;
newnode(0);
newnode(-1);
last = 0;
n = 0;
s[n] = -1;
fail[0] = 1;
}
int get_fail(int x){
while(s[n-len[x]-1] != s[n]) x = fail[x];
return x;
}
void add(int c){
c = c-'a';
s[++n] = c;
int cur = get_fail(last);
if(!next[cur][c]){
int now = newnode(len[cur]+2); //注意字符串长度为1的是由-1拓展过来的，要特判
fail[now] = next[get_fail(fail[cur])][c];
next[cur][c] = now;
num[now] = num[fail[now]] + 1;
}
last = next[cur][c];
cnt[last]++;
}
void count(){
for(int i=p-1;i>=0;i--)
cnt[fail[i]] += cnt[i];
}
}pam;
char s[maxn];
int main(){
scanf("%s",s);
int len = strlen(s);
pam.init();
for(int i=0;i<len;i++)
{
pam.add(s[i]);
}
pam.count();
long long ret = 0;
for(int i=2;i<pam.p;i++) //遍历所有的回文串
{
ret = max((long long)pam.len[i]*pam.cnt[i],ret);
}
cout<<ret<<endl;
return 0;
}
遍历
对于 A 中的每个回文子串，B 中和该子串相同的子串个数的总和
ll dfs(int an,int bn){
ll ret = 0;
for(int i=0;i<ALP;i++) if(pam1.next[an][i]!=0 && pam2.next[bn][i]!=0)
ret += (ll)pam1.cnt[pam1.next[an][i]] * pam2.cnt[pam2.next[bn][i]]
+ dfs(pam1.next[an][i],pam2.next[bn][i]);
return ret;
}
ll ret = dfs(0,0) + dfs(1,1);
最小表示法
给出一个循环字符串，输出最小字典序的开始下标
const int MAXN=100000+7;
char s[MAXN];
int n;
int get_minstring()
{
int len=n;
int i=0,j=1,k=0;
while(i<len&&j<len&&k<len)
{
int t=s[(i+k)%len]-s[(j+k)%len];//t值为两个字符比较的结果，而(i+k)%len是因为i+k会超过len又因为是循环同构串所以要%len。同理(j+k)%len也是一样的道理。
if(t==0)
k++;
else
{
if(t>0)
i+=k+1;
else
j+=k+1;
if(i==j)
j++;
k=0;
}
}
return min(i,j);
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%d%s",&n,s);
printf("%d
",get_minstring());
}
}
图论
k短路
Dijkstra
struct Edge
{
int to;
int value;
int next;
bool operator < (const Edge &t) const
{
return t.value < value;
}
};
struct Edge1
{
int to;
int value;
int next;
};
struct Node
{
int to;
int f, g;
bool operator < (const Node &t) const
{
if(t.f==f)
return t.g < g;
return t.f < f;
}
};
Edge edge[100010]; //存储边与原图反向的图
Edge1 edge1[100010]; //存储原图
int adj[1010], adj1[1010], visited[1010], dis[1010], edgeNum, edgeNum1;
int N, M;
void Dijkstra(int start)
{
int k;
Edge t, cur;
priority_queue<Edge> PQ;
for(k=0; k<N; k++)
{
visited[k] = 0;
dis[k] = INT_MAX;
}
t.to = start; //起始顶点
t.next = -1;
t.value = 0;
dis[start] = 0; //自己到自己路径为0
PQ.push(t);
visited[start] = 1; //标记以入队
while(!PQ.empty())
{
cur = PQ.top(); //出队
PQ.pop();
visited[cur.to] = 0; //标记出队
for(int tmp = adj[cur.to]; tmp != -1; tmp = edge[tmp].next)
{
if(dis[edge[tmp].to] > dis[cur.to] + edge[tmp].value)
{
dis[edge[tmp].to] = dis[cur.to] + edge[tmp].value;
if(visited[edge[tmp].to] == 0)
{
PQ.push(edge[tmp]);
visited[edge[tmp].to] = 1;
}
}
}
}
}
int A_star(int start, int end, int k)
{
Node e, ne;
int cnt = 0;
priority_queue<Node> PQ;
if(start==end)
k++;
if(dis[start]==INT_MAX) //无法到达终点
return -1;
e.to = start;
e.g = 0;
e.f = e.g + dis[e.to];
PQ.push(e);
while(!PQ.empty())
{
e = PQ.top();
PQ.pop();
if(e.to==end)
cnt++; //第cnt短路
if(cnt==k)
return e.g;
for(int i=adj1[e.to]; i!=-1; i=edge1[i].next)
{
ne.to = edge1[i].to;
ne.g = e.g + edge1[i].value;
ne.f = ne.g + dis[ne.to];
PQ.push(ne);
}
}
return -1;
}
void addEdge(int a, int b, int len) //反向图添加边
{
edge[edgeNum].to = b;
edge[edgeNum].next = adj[a];
edge[edgeNum].value = len;
adj[a] = edgeNum++;
}
void addEdge1(int a, int b, int len) //原图添加边
{
edge1[edgeNum1].to = b;
edge1[edgeNum1].next = adj1[a];
edge1[edgeNum1].value = len;
adj1[a] = edgeNum1++;
}
int main()
{
int a, b, len, i, s, t, k;
while(scanf("%d %d", &N, &M)!=EOF)
{
for(i=0; i<N; i++)
{
adj[i] = -1;
adj1[i] = -1;
}
for(edgeNum=i=0; i<M; i++)
{
scanf("%d %d %d", &a, &b, &len);
addEdge1(a-1, b-1, len); //构造原图
addEdge(b-1, a-1, len); //构造反向图
}
scanf("%d %d %d", &s, &t, &k);
Dijkstra(t-1); //求原图中各点到终点的最短路
int ans = A_star(s-1, t-1, k); //求第k短路
printf("%d
", ans);
}
return 0;
}
SPFA
#define INF 0x3f3f3f
#define maxn 1010
typedef pair<int, int> pii;
struct edge{int to, cost;};
vector<edge> G[maxn];//正向图，邻接表
vector<edge> rG[maxn];//反向图，邻接表
int s, t, k;
int n, m, ok;
int d[maxn], In_que[maxn];//反向spfa
void spfa(int s){
memset(In_que, 0, sizeof(In_que));
memset(d, INF, sizeof(d));
queue<int> Q;
Q.push(s);
d[s] = 0;
In_que[s] = 1;
while (!Q.empty()){
int u = Q.front();
Q.pop();
In_que[u] = 0;
for (int i = 0; i < rG[u].size(); i++){
edge e = rG[u][i];
int v = e.to, dd = d[u] + e.cost;
if (d[v] > dd){
d[v] = dd;
if (!In_que[v]){
Q.push(v);
In_que[v] = 1;
}
}
}
}
}
struct state{
int u, g, h;
bool operator < (const state &b) const{
return g + h > b.g + b.h;
}
state(int a, int b, int c):u(a), g(b), h(c){}
};
int cnt[maxn], kd[maxn];
void Astar(){
memset(cnt, 0, sizeof(cnt));
priority_queue<state> Q;//open表
Q.push(state(s, 0, d[s]));
while (!Q.empty()){
state cur = Q.top();
Q.pop();
int u = cur.u;
cnt[u]++;
if (cnt[u] == k && u == t){//终点第k次出队
printf("%d
", cur.g);
ok = 1;
break;
}
if (cnt[u] > k)//如果出队次数大于k，不用拓展了，因为一个点的第k短距离肯定是由周围点的
//前k短距离更新得来
continue;
for (int i = 0; i < G[u].size(); i++){
edge e = G[u][i];
if (cnt[e.to] != k){//如果这个点不在closed表中
Q.push(state(e.to, cur.g + e.cost, d[e.to]));
}
}
}
}
int main()
{
scanf("%d%d", &n, &m);
for (int i = 0; i < m; i++){
int u, v, c;
scanf("%d%d%d", &u, &v, &c);
G[u].push_back((edge){v, c});
rG[v].push_back((edge){u, c});//反向建边
}
scanf("%d%d%d", &s, &t, &k);
if (s == t) k++;//题目特殊要求。。
spfa(t);
Astar();
if (!ok)//如果没有第k短路
printf("-1
");
return 0;
}
生成树
瓶颈生成树
每次询问，在这两个点间找一条路径使得这个路径上最大的边权最小
#define N 50010
#define M 100010
struct Edge {
int from, to, dis, nex;
}edge[M<<1], hehe[M<<1];
bool cmp(Edge a, Edge b){return a.dis < b.dis;}
int head[N], edgenum;
void add(int u, int v, int d){
Edge E = {u,v,d,head[u]};
edge[edgenum] = E;
head[u] = edgenum++;
}
int f[N];
int find(int x){return x==f[x] ? x : f[x] = find(f[x]);}
bool Union(int x, int y){
int fx = find(x), fy = find(y);
if(fx == fy)return false;
if(fx>fy)swap(fx,fy);
f[fx] = fy;
return true;
}
int fa[N][20], dep[N], cost[N][20]; //cost[u][i]表示u点到u的第2^i次父亲 路径上最大边权
void bfs(int root){
queue<int>q;
fa[root][0] = root; dep[root] = 0;
cost[root][0] = 0;
q.push(root);
while(!q.empty()) {
int u = q.front(); q.pop();
for(int i = 1; i < 20; i++) {
fa[u][i] = fa[fa[u][i-1]][i-1];
cost[u][i] = max(cost[u][i-1], cost[fa[u][i-1]][i-1]);
}
for(int i = head[u]; ~i; i = edge[i].nex) {
int v = edge[i].to; if(v == fa[u][0])continue;
dep[v] = dep[u]+1; fa[v][0] = u; cost[v][0] = edge[i].dis;
q.push(v);
}
}
}
int work(int x, int y){
int ans = 0;
if(dep[x] < dep[y]) swap(x, y);
for(int i = 0; i < 20; i++) if((dep[x]-dep[y])&(1<<i)) {
ans = max(ans, cost[x][i]);
x = fa[x][i];
}
if(x == y)return ans;
for(int i = 19; i >= 0; i--)if(fa[x][i]!=fa[y][i]) {
ans = max(ans, max(cost[x][i], cost[y][i]));
x = fa[x][i], y = fa[y][i];
}
ans = max(ans, cost[x][0]);
ans = max(ans, cost[y][0]);
return ans;
}
int main(){
int i, j, u, v, d, query, n, m, fir = 0;
while(~scanf("%d %d",&n,&m)){
if(fir++)puts("");
memset(head, -1, sizeof head); edgenum = 0;
for(i = 1; i <= n; i++) f[i] = i;
for(i = 0; i < m; i++){
read(hehe[i].from); read(hehe[i].to); read(hehe[i].dis);
}
sort(hehe, hehe + m, cmp);
int siz = 0;
for(int i = 0; i < m && siz < n; i++) {
if(Union(hehe[i].from, hehe[i].to)){
siz++;
add(hehe[i].from, hehe[i].to, hehe[i].dis);
add(hehe[i].to, hehe[i].from, hehe[i].dis);
}
}
bfs(1);
read(query);
while(query--){
read(u); read(v);
printf("%d
", work(u, v));
}
}
return 0;
}
k小生成树
输出（是k小生成树的权值和）
const int MAXN = 1000+100;
const int M = 3000;
const int MO=(1LL<<32);
int n,m,k,ca,cb;
struct node{
int v,w,nex;
}e[M*2];
int EN,head[M*2],fa[MAXN],A[100010],B[100010];
struct edge{
int u,v,w;
edge(int u,int v,int w):u(u),v(v),w(w){}
edge(){}
};
vector<edge> ve;
void init()
{
memset(head,-1,sizeof head);
EN = 0;
ve.clear();
ca=0;cb=0;
for(int i=0;i<=n;i++)fa[i]=i;
}
void add(int u,int v,int w)
{
e[EN].v=v,e[EN].w=w,e[EN].nex=head[u];
head[u]=EN++;
}
int fand(int x)
{
return fa[x]==x?x:fa[x]=fand(fa[x]);
}
bool dfs(int u,int f,int en)
{
if(u==en)return true;
for(int i=head[u];~i;i=e[i].nex)
{
if(i==f) continue;
if(dfs(e[i].v,i^1,en))
{
B[cb++]=e[i].w;
return true;
}
}
return false;
}
bool cmp(int a,int b)
{
return a>b;
}
void merg()
{
if(ca > cb) swap(ca,cb),swap(A,B);
int siz = min(k,ca*cb);
priority_queue<pair<int,int> >q;
for(int i=0;i<ca;i++) q.push(make_pair(A[i]+B[0],0));
ca=siz;
for(int i=0;i<siz;i++)
{
pair<int,int>p = q.top();q.pop();
A[i] = p.first;
if(p.second+1<cb) q.push(make_pair(p.first-B[p.second]+B[p.second+1],p.second+1));
}
}
main()
{
ios::sync_with_stdio(false);
int tt=0;
while(cin>>n>>m)
{
init();
int sum=0;
int u,v,w;
while(m--)
{
cin>>u>>v>>w;
if(fand(u)!=fand(v))
{
add(u,v,w);
add(v,u,w);
fa[fand(u)]=fand(v);
}else
{
ve.push_back(edge(u,v,w));
}
sum+=w;
}cin>>k;
for(auto E : ve)
{
cb=0;
B[cb++]=E.w;
dfs(E.u,-1,E.v);
sort(B,B+cb,cmp);
if(ca==0)
{
for(int i=0;i<cb;i++)
A[i]=B[i];
ca=cb;
}else
merg();
}
int ans=0;
for(int i=0;i<ca;i++)
{
ans+=(sum-A[i])*(i+1);
ans%=MO;
}
if(ans==0) ans=sum;
ans%=MO;
cout<<"Case #"<<++tt<<": "<<ans<<endl;
}
}
虚树
#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 1e6+10;
#define FOR(i,L,R) for(register int i=(L);i<=(R);++i)
#define REP(j,R,L) for(register int j=(R);j>=(L);--j)
int n;
struct Graph{//处理最初图，求lca
vector<int>G[N];
int cnt,xu[N],dep[N],p[N][27];
inline int lca(int x,int y){
if(dep[x]<dep[y])swap(x,y);
REP(i,log2(dep[x]),0)if(dep[x]-(1<<i)>=dep[y])x=p[x][i];
if(x==y)return y;
REP(i,log2(dep[x]),0)if(p[x][i]^p[y][i])x=p[x][i],y=p[y][i];
return p[x][0];
}
inline void ST(){
int k=log2(n);
FOR(j,1,k)FOR(i,1,n)if(~p[i][j-1])p[i][j]=p[p[i][j-1]][j-1];
}
inline void dfs(int u,int fa,int depth){
xu[u]=++cnt;
int sz=G[u].size()-1;
FOR(i,0,sz)if(G[u][i]^fa)dfs(G[u][i],p[G[u][i]][0]=u,dep[G[u][i]]=depth+1);
}
inline void init(){
memset(p,-1,sizeof(p));
scanf("%d",&n);
FOR(i,1,n-1){
int x,y;scanf("%d%d",&x,&y);
G[x].push_back(y);
G[y].push_back(x);
}
}
}g;
bool cmp(int a,int b){
return g.xu[a]<g.xu[b];
}
struct Fake_Tree{//应该是这么翻译吧……
struct edge{
int to,next,w;
}a[N];//这里不用vector是因为清空很慢
int m,h[N],sig[N],flag[N];
int cnt,top,stk[N];
#define next a[p].next
#define to a[p].to
#define w a[p].w
inline void solv(){
……
FOR(i,1,m)flag[sig[i]]=0;//记得清空。。。
}
inline void add(int u,int v){
if(u==v)return ;//注意这里很重要！
a[++cnt]=(edge){v,h[u],g.dep[v]-g.dep[u]};h[u]=cnt;//因为每次维护一条树链且是单位长度，所以u和v的距离就是他们之间的深度
}
inline void init(){
scanf("%d",&m);
FOR(i,1,m)scanf("%d",sig+i);//读入key point
sort(sig+1,sig+m+1,cmp);
FOR(i,1,m)flag[sig[i]]=1;//标记
//注意有的题可以删去一些关键点
cnt=top=0;
stk[++top]=1;
FOR(i,1,m){
int now=sig[i],lca=g.lca(now,stk[top]);
while(g.dep[lca]<g.dep[stk[top-1]]){//因为是维护树链，所以可用深度判断
add(stk[top-1],stk[top]);
--top;
}add(lca,stk[top]);
stk[top]=lca;//直接覆盖，注意上面使用top-1判断
if(stk[top]!=now)stk[++top]=now;//判断稳妥一点
}while(--top)add(stk[top],stk[top+1]);//最后剩在栈里的一条链
}
}z;
int main(){
g.init(); //读入原图
g.dfs(1,g.p[1][0]=0,g.dep[1]=1);
g.ST();
int Case;scanf("%d",&Case);
while(Case--){
z.init();//读入关键点
z.solv();
}return 0;
}
欧拉回路
第一行一个整数 t，表示子任务编号。t∈{1,2}，如果 t=1 则表示处理无向图的情况，如果t=2 则表示处理有向图的情况。
第二行两个整数 n,m，表示图的结点数和边数。
接下来 m 行中，第 i 行两个整数 vi,ui，表示第 i 条边（从 1 开始编号）。保证 1≤vi,ui≤n。
1.如果 t=1 则表示 vi 到 ui 有一条无向边。
2.如果 t=2 则表示 vi 到 ui 有一条有向边。
图中可能有重边也可能有自环。
int T,n,m,p,d[100010],ru[100010],chu[100010],ans[200010],h[100010];
vector<int> b;
struct data
{
int x,id;
data(int a=0,int b=0):x(a),id(b){}
};
vector<data> a[100010];
bool v[200010];
void dfs1(int x,int y)
{
if (a[x].size()>h[x])
{
while (v[abs(a[x][h[x]].id)])
{
h[x]++;
if (h[x]==a[x].size()) break;
}
if (a[x].size()>h[x])
{
int i=a[x][h[x]].x,j=a[x][h[x]].id;
h[x]++;v[abs(j)]=1;
if (i!=y) dfs1(i,y);
ans[++p]=j;
}
}
if (a[x].size()>h[x])
{
while (v[abs(a[x][h[x]].id)])
{
h[x]++;
if (h[x]==a[x].size()) break;
}
if (a[x].size()>h[x]) dfs1(x,x);
}
}
void dfs2(int x,int y)
{
if (a[x].size()>h[x])
{
int i=a[x][h[x]].x,j=a[x][h[x]].id;
h[x]++;
if (i!=y) dfs2(i,y);
ans[++p]=j;
}
if (a[x].size()>h[x]) dfs2(x,x);
}
int main()
{
read(T);
if (T==1)
{
read(n);read(m);
int x,y;
for (int i=1;i<=m;i++)
read(x),read(y),d[x]++,d[y]++,a[x].push_back(data(y,i)),a[y].push_back(data(x,-i));
x=0;
for (int i=1;i<=n;i++)
if (d[i]%2==1) {puts("NO");return 0;}
else if (d[i]>0) x=i;
memset(h,0,sizeof(h));
p=0;dfs1(x,x);
if (p<m) {puts("NO");return 0;}
puts("YES");
for (int i=m;i>1;i--)
print(ans[i]),putchar(' ');
if (m>0) print(ans[1]),puts("");
}
else if (T==2)
{
read(n);read(m);
int x,y;
for (int i=1;i<=m;i++)
read(x),read(y),chu[x]++,ru[y]++,a[x].push_back(data(y,i));
x=0;
for (int i=1;i<=n;i++)
if (ru[i]!=chu[i]) {puts("NO");return 0;}
else if (ru[i]>0) x=i;
memset(h,0,sizeof(h));
p=0;dfs2(x,x);
if (p<m) {puts("NO");return 0;}
puts("YES");
for (int i=m;i>1;i--)
print(ans[i]),putchar(' ');
if (m>0) print(ans[1]),puts("");
}
return 0;
}