summary：57
bzoj1741：裸二分图最大匹配
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
#define qwq(x) for(edge *o=head[x];o;o=o->next)
int read(){
int x=0;char c=getchar();
while(!isdigit(c)) c=getchar();
while(isdigit(c)) x=x*10+c-'0',c=getchar();
return x;
}
const int nmax=1005;
const int maxn=10005;
struct edge{
int to;edge *next;
};
edge es[maxn],*pt=es,*head[nmax];
bool vis[nmax];int match[nmax];
void add(int u,int v){
pt->to=v;pt->next=head[u];head[u]=pt++;
}
bool dfs(int x){
qwq(x) if(!vis[o->to]){
vis[o->to]=1;
if(!match[o->to]||dfs(match[o->to])){
match[o->to]=x;return true;
}
}
return false;
}
int main(){
int n=read(),m=read(),u,v;
rep(i,1,m) u=read(),v=read(),add(u,v+n);
int ans=0;
rep(i,1,n){
clr(vis,0);
if(dfs(i)) ans++;
}
printf("%d
",ans);
return 0;
}
bzoj1742：dp题。我想的总是点的情况，比如到当前是那一步用步数来转移还是不行的。需要考虑的转换情况的这段区间的情况，不要局限于死思维！！！ 其实跟以前做过的差不多，枚举区间的长短。为什么想不出来！！！
dp[j][0]改变了居然样例能过。。。然后就WA了。
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
int read(){
int x=0;char c=getchar();
while(!isdigit(c)) c=getchar();
while(isdigit(c)) x=x*10+c-'0',c=getchar();
return x;
}
const int nmax=1005;
int dp[nmax][2],a[nmax];
int main(){
int n=read(),s=read();
rep(i,1,n) a[i]=read();
sort(a+1,a+n+1);
rep(i,1,n) dp[i][0]=dp[i][1]=abs(s-a[i])*n;
rep(i,2,n){
for(int j=1;j+i-1<=n;j++){
int k=j+i-1,tmp=dp[j][0];
dp[j][0]=min(dp[j+1][0]+(a[j+1]-a[j])*(n-i+1),dp[j+1][1]+(a[k]-a[j])*(n-i+1));
dp[j][1]=min(dp[j][1]+(a[k]-a[k-1])*(n-i+1),tmp+(a[k]-a[j])*(n-i+1));
}
}
printf("%d
",min(dp[1][0],dp[1][1]));
return 0;
}
bzoj1750：dp。。。每个点都可以转或者不转。。。
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
const int nmax=1005;
const int maxn=35;
int dp[nmax][maxn],a[nmax];
void maxs(int &a,int b){
if(a<b) a=b;
}
int main(){
int n,m;scanf("%d%d",&n,&m);
rep(i,1,n) scanf("%d",&a[i]);
rep(i,1,n){
rep(j,0,i){
if(j>m) break;
if(j) dp[i][j]=max(dp[i-1][j],dp[i-1][j-1])+(j%2+1==a[i]);
else dp[i][j]=dp[i-1][j]+(j%2+1==a[i]);
}
}
rep(i,1,m) maxs(dp[n][i],dp[n][i-1]);
printf("%d
",dp[n][m]);
return 0;
}
bzoj1752：最短路。。。我。。。
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
#define qwq(x) for(edge *o=head[x];o;o=o->next)
int read(){
int x=0;char c=getchar();
while(!isdigit(c)) c=getchar();
while(isdigit(c)) x=x*10+c-'0',c=getchar();
return x;
}
const int nmax=1005;
const int maxn=100005;
const int inf=0x7f7f7f7f;
struct edge{
int to,dist;edge *next;
};
edge es[maxn],*pt=es,*head[nmax];
void add(int u,int v,int d){
pt->to=v;pt->dist=d;pt->next=head[u];head[u]=pt++;
pt->to=u;pt->dist=d;pt->next=head[v];head[v]=pt++;
}
int dist[nmax];
struct node{
int x,dist;
node(int x,int dist):x(x),dist(dist){};
bool operator<(const node&rhs)const{
return dist>rhs.dist;}
};
priority_queue<node>q;
void dijkstra(){
clr(dist,0x7f);dist[1]=0;q.push(node(1,0));
while(!q.empty()){
node o=q.top();q.pop();
int tx=o.x,td=o.dist;
if(dist[tx]!=td) continue;
qwq(tx) if(dist[o->to]>td+o->dist){
dist[o->to]=td+o->dist;q.push(node(o->to,dist[o->to]));
}
}
}
int main(){
int m=read(),n=read(),u,v,d;
rep(i,1,m) u=read(),v=read(),d=read(),add(u,v,d);
dijkstra();
printf("%d
",dist[n]);
return 0;
}
bzoj1753：我、、、
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
int read(){
int x=0;char c=getchar();
while(!isdigit(c)) c=getchar();
while(isdigit(c)) x=x*10+c-'0',c=getchar();
return x;
}
int a[1000005];
int main(){
int n=read();
rep(i,1,n) a[i]=read();
sort(a+1,a+n+1);
printf("%d
",a[(n+1)>>1]);
return 0;
}
bzoj1754：高精度乘法。。。
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
int a[45],b[45],ans[100];
char s[45],t[45];
int main(){
scanf("%s%s",s+1,t+1);
int lena=strlen(s+1),lenb=strlen(t+1),len=lena+lenb;
rep(i,1,lena) a[i]=s[lena-i+1]-'0';
rep(i,1,lenb) b[i]=t[lenb-i+1]-'0';
rep(i,1,lena) rep(j,1,lenb){
ans[i+j-1]+=a[i]*b[j];
if(ans[i+j-1]>9) {
ans[i+j]+=ans[i+j-1]/10;
ans[i+j-1]%=10;
}
}
if(ans[lenb+lena-1]>9) {
ans[lena+lenb]=ans[lena+lenb-1]/10;
ans[lena+lenb-1]%=10;
len++;
}
while(ans[len]==0) len--;
dwn(i,len,1) printf("%d",ans[i]);printf("
");
return 0;
}
bzoj1755：我。。。
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
int main(){
double a,b;int n;
scanf("%lf%lf%d",&a,&b,&n);
a/=100;a+=1;
rep(i,1,n) b*=a;
int ans=(int)b;
printf("%d
",ans);
return 0;
}
bzoj1770：高斯消元然后爆搜自由元以前写过。。。
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<bitset>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
int read(){
int x=0;char c=getchar();
while(!isdigit(c)) c=getchar();
while(isdigit(c)) x=x*10+c-'0',c=getchar();
return x;
}
const int nmax=40;
const int inf=0x7f7f7f7f;
bitset<nmax>a[nmax];
void gauss(int n,int m){
rep(i,1,n){
int j;for(j=i;j<=n&&!a[j][i];j++)
if(j<n) continue;
if(j!=i) swap(a[i],a[j]);
rep(j,i+1,n) if(a[j][i]) a[j]^=a[i];
}
}
int res=inf,ans[nmax],tot=0;
int n,m;
void mins(int &a,int b){
if(a>b) a=b;
}
void dfs(int x){
if(tot>=res) return ;
if(!x) {
mins(res,tot);return ;
}
if(a[x][x]){
int t=a[x][n+1];
rep(i,x+1,n) if(a[x][i]) t^=ans[i];
ans[x]=t;
if(t) tot++;
dfs(x-1);
if(t) tot--;
}else{
ans[x]=0;dfs(x-1);
ans[x]=1;tot++;dfs(x-1);tot--;
}
}
int main(){
n=read(),m=read();int u,v;
rep(i,1,n) a[i][i]=a[i][n+1]=1;
rep(i,1,m) u=read(),v=read(),a[u][v]=a[v][u]=1;
gauss(n,n+1);dfs(n);
printf("%d
",res);
return 0;
}
洛谷3793：输出第n组勾股数。不会。。。在网上找勾股数的规律被我找到了。。。
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(ll i=s;i<=t;i++)
#define ll long long
int main(){
ll n;
while(scanf("%lld",&n)==1){
n+=2;
if(n%2){
printf("%lld %lld %lld
",n,(n*n-1)/2,(n*n+1)/2);
}else printf("%lld %lld %lld
",n,(n/2)*(n/2)-1,(n/2)*(n/2)+1);
}
return 0;
}
洛谷3832：田忌赛马。不会。。。神贪心！
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
int read(){
int x=0,f=1;char c=getchar();
while(!isdigit(c)) {
if(c=='-') f=false;c=getchar();
}
while(isdigit(c)) x=x*10+c-'0',c=getchar();
return x*f;
}
const int nmax=5005;
int a[nmax],b[nmax];
int main(){
int n=read();
rep(i,1,n) a[i]=read();
rep(i,1,n) b[i]=read();
sort(a+1,a+n+1);sort(b+1,b+n+1);
int l=1,s=1,r=n,t=n,ans=0;
while(l<=r){
if(a[l]>b[s]) ans+=200,l++,s++;
else if(a[r]>b[t]) ans+=200,r--,t--;
else ans+=(a[l]!=b[t])*(-200),++l,--t;
}
printf("%d
",ans);
return 0;
}
bzoj1774：将点权排序后用floyed就好了。没有考虑重边的情况wa了一次。看清楚题！！！
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
int read(){
int x=0;char c=getchar();
while(!isdigit(c)) c=getchar();
while(isdigit(c)) x=x*10+c-'0',c=getchar();
return x;
}
const int nmax=255;
int dist[nmax][nmax],ans[nmax][nmax],w[nmax];
struct node{
int w,id;
bool operator<(const node&rhs)const{
return w<rhs.w;}
};
node ns[nmax];
int main(){
int n=read(),m=read(),q=read(),u,v,d;
rep(i,1,n) w[i]=ns[i].w=read(),ns[i].id=i;
sort(ns+1,ns+n+1);
clr(dist,0x3f);clr(ans,0x3f);
rep(i,1,n) dist[i][i]=ans[i][i]=0;
rep(i,1,m) u=read(),v=read(),d=read(),dist[u][v]=dist[v][u]=min(dist[u][v],d);
rep(tk,1,n){
int k=ns[tk].id;
rep(i,1,n) rep(j,1,n) {
dist[i][j]=min(dist[i][j],dist[i][k]+dist[k][j]);
ans[i][j]=min(ans[i][j],dist[i][j]+max(ns[tk].w,max(w[i],w[j])));
}
}
rep(i,1,q) u=read(),v=read(),printf("%d
",ans[u][v]);
return 0;
}
bzoj1782：树链剖分O(nlogn)还是可以写的。看了一下题解可以用类似dfs序+bit解决。子树消掉+bit的经典应用。
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
#define qwq(x) for(edge *o=head[x];o;o=o->next)
int read(){
int x=0;char c=getchar();
while(!isdigit(c)) c=getchar();
while(isdigit(c)) x=x*10+c-'0',c=getchar();
return x;
}
const int nmax=100005;
const int inf=0x7fffffff;
int sum[nmax<<2],n,id[nmax],ans[nmax];
struct edge{
int to;edge *next;
};
edge es[nmax<<1],*pt=es,*head[nmax];
void add(int u,int v){
pt->to=v;pt->next=head[u];head[u]=pt++;
pt->to=u;pt->next=head[v];head[v]=pt++;
}
void update(int x,int w){
for(int i=x;i<=n;i+=(i&-i)) sum[i]+=w;
}
int query(int x){
int ans=0;
for(int i=x;i;i-=(i&-i)) ans+=sum[i];
return ans;
}
void dfs(int x,int fa){
int t=id[x];
ans[t]=query(id[x]);update(id[x],1);
qwq(x) if(o->to!=fa) dfs(o->to,x);
update(id[x],-1);
}
int main(){
n=read();int u,v;
rep(i,1,n-1) u=read(),v=read(),add(u,v);
rep(i,1,n) u=read(),id[u]=i;
dfs(1,0);
rep(i,1,n) printf("%d
",ans[i]);
return 0;
}
bzoj1827：树形dp。想处理子树的情况，再利用父结点处理非子树的情况就可以了。
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
#define qwq(x) for(edge *o=head[x];o;o=o->next)
#define ll long long
int read(){
int x=0;char c=getchar();
while(!isdigit(c)) c=getchar();
while(isdigit(c)) x=x*10+c-'0',c=getchar();
return x;
}
const int nmax=100005;
const ll inf=1e18;
ll w[nmax],size[nmax];
ll f[nmax],g[nmax],n,sum=0;
struct edge{
int to;ll dist;edge *next;
};
edge es[nmax<<1],*pt=es,*head[nmax];
void add(int u,int v,ll d){
pt->to=v;pt->dist=d;pt->next=head[u];head[u]=pt++;
pt->to=u;pt->dist=d;pt->next=head[v];head[v]=pt++;
}
void dfs(int x,int fa){
size[x]=w[x];
qwq(x) if(o->to!=fa){
dfs(o->to,x);size[x]+=size[o->to];
f[x]+=f[o->to]+size[o->to]*o->dist;
}
}
void DFS(int x,int fa){
qwq(x) if(o->to!=fa){
g[o->to]=f[o->to]+g[x]-f[o->to]-size[o->to]*o->dist+(sum-size[o->to])*o->dist;
DFS(o->to,x);
}
}
int mins(ll &a,ll b){
if(a>b) a=b;
}
int main(){
n=read();int u,v;ll d;
rep(i,1,n) w[i]=read(),sum+=w[i];
rep(i,1,n-1) u=read(),v=read(),d=read(),add(u,v,d);
dfs(1,0);
g[1]=f[1];
DFS(1,0);
ll ans=inf;
rep(i,1,n) mins(ans,g[i]);
printf("%lld
",ans);
return 0;
}
bzoj3196：看题目肯定是二分答案。那么二分答案后贪心判断！
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
int read(){
int x=0;char c=getchar();
while(!isdigit(c)) c=getchar();
while(isdigit(c)) x=x*10+c-'0',c=getchar();
return x;
}
const int nmax=505;
const int inf=0x7fffffff;
int map[nmax][nmax],n,m,A,B;
int check(int s,int t,int x){
int cur=1,sum,ans=0;bool flag;
while(1){
sum=0;flag=false;
rep(i,cur,m){
sum+=map[t][i]-map[s][i];
if(sum>=x) {
flag=true;ans++;cur=i+1;break;
}
}
if(cur>m||!flag) break;
}
return ans;
}
bool pd(int x){
int cur=1,ans=0;bool flag;
while(1){
flag=false;
rep(i,cur,n) {
if(check(cur-1,i,x)>=B){
flag=true;ans++;cur=i+1;break;
}
}
if(cur>n||!flag) break;
}
return ans>=A;
}
int main(){
n=read(),m=read(),A=read(),B=read();
rep(i,1,n) rep(j,1,m) map[i][j]=map[i-1][j]+read();
int l=0,r=inf,mid,ans=0;
while(l<=r){
mid=(l+r)>>1;
if(pd(mid)) ans=mid,l=mid+1;
else r=mid-1;
}
printf("%d
",ans);
return 0;
}
bzoj2060：树形dp
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
#define qwq(x) for(edge *o=head[x];o;o=o->next)
int read(){
int x=0;char c=getchar();
while(!isdigit(c)) c=getchar();
while(isdigit(c)) x=x*10+c-'0',c=getchar();
return x;
}
const int nmax=50005;
const int inf=0x7f7f7f7f;
int dp[nmax][2];
struct edge{
int to;edge *next;
};
edge es[nmax<<1],*pt=es,*head[nmax];
void add(int u,int v){
pt->to=v;pt->next=head[u];head[u]=pt++;
pt->to=u;pt->next=head[v];head[v]=pt++;
}
void dfs(int x,int fa){
dp[x][1]=1;
qwq(x) if(o->to!=fa){
int to=o->to;dfs(to,x);
dp[x][0]+=max(dp[to][0],dp[to][1]);
dp[x][1]+=dp[to][0];
}
}
int main(){
int n=read(),u,v;
rep(i,1,n-1) u=read(),v=read(),add(u,v);
dfs(1,0);
printf("%d
",max(dp[1][1],dp[1][0]));
return 0;
}
bzoj2274：dp.f[i]+=f[j],sum[i]-sum[j]>=0;O(n2) orz又跪了 =>加树状数组搞一下就行了！！！（对bit有点了解了。。。
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
int read(){
int x=0,f=1;char c=getchar();
while(!isdigit(c)) {
if(c=='-') f=-1;c=getchar();
}
while(isdigit(c)) x=x*10+c-'0',c=getchar();
return x*f;
}
const int nmax=100005;
const int mod=1000000009;
int a[nmax],b[nmax],sum[nmax<<2],dp[nmax],n,N;
int query(int x){
int ans=0;
for(int i=x;i;i-=(i&-i)) {
ans=(ans+sum[i])%mod;
}
return ans;
}
void insert(int x,int w){
for(int i=x;i<=N;i+=(i&-i)) sum[i]=(sum[i]+w)%mod;
}
int main(){
n=read();int u;
rep(i,1,n) u=read(),b[i]=b[i-1]+u,a[i]=b[i];
b[0]=0;sort(b,b+n+1);
N=unique(b,b+n+1)-b;
u=lower_bound(b,b+N,0)-b;
insert(++u,1);
rep(i,1,n) {
u=lower_bound(b,b+N,a[i])-b;
dp[i]=query(++u);
insert(u,dp[i]);
}
printf("%d
",dp[n]);
return 0;
}
/*
5 -1 -1 -1 -1 -1*/
bzoj2442：开始的时候想了贪心无解。。然后dp！。嘛o(nk)的还是很容易的。。。学到了单调队列优化dp。。而且连dp[i][0],dp[i][1]都可以省略了。用了fread果然快了挺多的。。。好神啊啊啊 ps：滑块不就用单调队列么2333
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;++i)
#define dwn(i,s,t) for(int i=s;i>=t;--i)
#define clr(x,c) memset(x,c,sizeof(x))
#define ll long long
char buf[1000005],*ptr=buf-1;
ll read(){
ll x=0;char c=*++ptr;
while(!isdigit(c)) c=*++ptr;
while(isdigit(c)) x=x*10+c-'0',c=*++ptr;
return x;
}
const int nmax=100005;
ll dp[nmax],sum[nmax],q[nmax];
int main(){
fread(buf,1,1000000,stdin);
ll n=read(),k=read();
rep(i,1,n) sum[i]=sum[i-1]+read();
int l=1,r=1;
rep(i,1,n){
while(l<=r&&q[l]<i-k) ++l;
if(i<=k) dp[i]=sum[i];
else dp[i]=dp[q[l]-1]+sum[i]-sum[q[l]];
while(l<=r&&dp[q[r]-1]-sum[q[r]]<dp[i-1]-sum[i]) --r;
q[++r]=i;
}
printf("%lld
",dp[n]);
return 0;
}
bzoj2591：嘛是树形dp嘛。。。不过居然写了很久！！！没有将细节考虑清楚前不要写！写到一半发现有误停下来全部想清楚再写！！！不要这个改一下那个改一下！！！ 哦树形dp还是先处理子树的情况然后再有父结点弄出非子树的情况。
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
#define qwq(x) for(edge *o=head[x];o;o=o->next)
int read(){
int x=0;char c=getchar();
while(!isdigit(c)) c=getchar();
while(isdigit(c)) x=x*10+c-'0',c=getchar();
return x;
}
const int nmax=100005;
const int inf=0x7f7f7f7f;
int f[nmax][25],g[nmax][25],w[nmax],n,K;
struct edge{
int to;edge *next;
};
edge es[nmax<<1],*pt=es,*head[nmax];
void add(int u,int v){
pt->to=v;pt->next=head[u];head[u]=pt++;
pt->to=u;pt->next=head[v];head[v]=pt++;
}
void dfs(int x,int fa){
f[x][0]=w[x];
qwq(x) if(o->to!=fa){
dfs(o->to,x);
rep(i,1,K) f[x][i]+=f[o->to][i-1];
}
}
void DFS(int x,int fa){
g[x][0]=w[x];g[x][1]=f[x][1]+w[fa];
rep(i,2,K) g[x][i]=f[x][i]+g[fa][i-1]-f[x][i-2];
qwq(x) if(o->to!=fa) DFS(o->to,x);
}
int main(){
n=read(),K=read();int u,v;
rep(i,1,n-1) u=read(),v=read(),add(u,v);
rep(i,1,n) w[i]=read();
dfs(1,0);
rep(i,0,K) g[1][i]=f[1][i];
qwq(1) DFS(o->to,1);
rep(i,1,n) {
rep(j,0,K) g[i][j]+=g[i][j-1];
printf("%d
",g[i][K]);
}
return 0;
}
bzoj3050：测了自己手造数据就直接交了。output limited error。。。注意注意注意中间输出要删掉中间输出要删掉中间输出要删掉！而且最好肉眼看一下有没有写错QAQ。 还有还有还有！有时候编译器会出现诡异错误按f9的时候直接整个程序就没了！所以我们要先保存好文件再开始写！
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
#define lson l,mid,x<<1
#define rson mid+1,r,x<<1|1
int read(){
int x=0;char c=getchar();
while(!isdigit(c)) c=getchar();
while(isdigit(c)) x=x*10+c-'0',c=getchar();
return x;
}
const int nmax=2000005;
const int inf=0x7f7f7f7f;
int lmax[nmax],omax[nmax],rmax[nmax],col[nmax];
char s[5];
int n;
void build(int l,int r,int x){
lmax[x]=rmax[x]=omax[x]=r-l+1;col[x]=-1;
if(l==r) return ;
int mid=(l+r)>>1;build(lson);build(rson);
}
void pushdown(int x,int cnt){
if(col[x]!=-1){
col[x<<1]=col[x<<1|1]=col[x];
lmax[x<<1]=rmax[x<<1]=omax[x<<1]=col[x]?0:cnt-(cnt>>1);
lmax[x<<1|1]=rmax[x<<1|1]=omax[x<<1|1]=col[x]?0:cnt>>1;
col[x]=-1;
}
}
void pushup(int x,int cnt){
lmax[x]=lmax[x<<1]==cnt-(cnt>>1)?lmax[x<<1]+lmax[x<<1|1]:lmax[x<<1];
rmax[x]=rmax[x<<1|1]==(cnt>>1)?rmax[x<<1|1]+rmax[x<<1]:rmax[x<<1|1];
omax[x]=max(rmax[x<<1]+lmax[x<<1|1],max(omax[x<<1],omax[x<<1|1]));
}
int query(int p,int l,int r,int x){
if(r-l+1==p) return l;
pushdown(x,r-l+1);
int mid=(l+r)>>1;
if(omax[x<<1]>=p) return query(p,lson);
if(rmax[x<<1]+lmax[x<<1|1]>=p) return mid-rmax[x<<1]+1;
return query(p,rson);
}
void update(int tl,int tr,int p,int l,int r,int x){
if(tl<=l&&tr>=r){
col[x]=p;
lmax[x]=rmax[x]=omax[x]=p?0:r-l+1;
return ;
}
pushdown(x,r-l+1);
int mid=(l+r)>>1;
if(tl<=mid) update(tl,tr,p,lson);
if(tr>mid) update(tl,tr,p,rson);
pushup(x,r-l+1);
}
int main(){
n=read();int m=read(),u,v,ans=0;
build(1,n,1);
rep(i,1,m) {
scanf("%s",s);
if(s[0]=='A'){
u=read();
if(omax[1]<u) ans++;
else {
v=query(u,1,n,1);
update(v,v+u-1,1,1,n,1);
}
}else{
u=read(),v=read();
update(u,v,0,1,n,1);
}
}
printf("%d
",ans);
return 0;
}
bzoj3407：背包dp。。。
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
int read(){
int x=0;char c=getchar();
while(!isdigit(c)) c=getchar();
while(isdigit(c)) x=x*10+c-'0',c=getchar();
return x;
}
int a[505],f[45005];
int main(){
int m=read(),n=read();
rep(i,1,n) a[i]=read();
f[0]=1;
rep(i,1,n) dwn(j,m,a[i]) f[j]|=f[j-a[i]];
dwn(i,m,0) if(f[i]){
printf("%d
",i);return 0;
}
return 0;
}
bzoj3445：删掉的肯定是最短路中的边。。。枚举一下就好了。。。
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
#define qwq(x) for(edge *o=head[x];o;o=o->next)
int read(){
int x=0;char c=getchar();
while(!isdigit(c)) c=getchar();
while(isdigit(c)) x=x*10+c-'0',c=getchar();
return x;
}
const int nmax=255;
const int maxn=250005;
struct edge{
int to,dist,id;edge *next;
};
edge es[maxn<<1],*pt=es,*head[nmax];
int n,m,dist[nmax],pre[nmax],fa[nmax];
void add(int u,int v,int d,int id){
pt->to=v;pt->dist=d;pt->id=id;pt->next=head[u];head[u]=pt++;
pt->to=u;pt->dist=d;pt->id=id;pt->next=head[v];head[v]=pt++;
}
struct node{
int x,dist;
node(int x,int dist):x(x),dist(dist){};
node(){};
bool operator<(const node&rhs)const{
return dist>rhs.dist;}
};
priority_queue<node>q;
void maxs(int &a,int b){
if(a<b) a=b;
}
void dijkstra(){
clr(dist,0x7f);dist[1]=0;q.push(node(1,0));
node o;int tx,td;
while(!q.empty()){
o=q.top();q.pop();
tx=o.x;td=o.dist;
if(dist[tx]!=td) continue;
qwq(tx) if(dist[o->to]>td+o->dist){
dist[o->to]=td+o->dist;pre[o->to]=tx;fa[o->to]=o->id;
q.push(node(o->to,dist[o->to]));
}
}
}
void DIJKSTRA(){
clr(dist,0x7f);dist[1]=0;q.push(node(1,0));
node o;int tx,td;
while(!q.empty()){
o=q.top();q.pop();
tx=o.x;td=o.dist;
if(dist[tx]!=td) continue;
qwq(tx) if(dist[o->to]>td+o->dist){
dist[o->to]=td+o->dist;
q.push(node(o->to,dist[o->to]));
}
}
}
int main(){
n=read(),m=read();
int u,v,d,tmp,temp;
rep(i,1,m) u=read(),v=read(),d=read(),add(u,v,d,i);
dijkstra();tmp=dist[n];
for(int i=n;i!=1;i=pre[i]) {
u=pre[i];
qwq(u) if(o->id==fa[i]) o->dist<<=1;
qwq(i) if(o->id==fa[i]) o->dist<<=1;
DIJKSTRA();maxs(temp,dist[n]);
qwq(u) if(o->id==fa[i]) o->dist>>=1;
qwq(i) if(o->id==fa[i]) o->dist>>=1;
}
printf("%d
",temp-tmp);
return 0;
}
bzoj3048:二分答案然后判断？k(n-k)最坏的情况下50000*50000. 嗯还是有暴力分的。sad又跪了=> 用类似单调队列的方法扫一遍就好了。。。
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
int read(){
int x=0;char c=getchar();
while(!isdigit(c)) c=getchar();
while(isdigit(c)) x=x*10+c-'0',c=getchar();
return x;
}
const int nmax=100005;
int a[nmax],b[nmax],cnt[nmax];
bool vis[nmax];
void maxs(int &a,int b){
if(a<b) a=b;
}
int main(){
int n=read(),k=read();
rep(i,1,n) a[i]=b[i]=read();
sort(b+1,b+n+1);
int m=unique(b+1,b+n+1)-b;
rep(i,1,n) a[i]=lower_bound(b+1,b+m+1,a[i])-b;
int l=1,r=1,sum=0,ans=1;clr(vis,0);
while(r<=n){
while(vis[a[r]]||(!vis[a[r]])&&sum<k+1){
if(vis[a[r]]) maxs(ans,++cnt[a[r]]);
else vis[a[r]]=1,++cnt[a[r]],sum++;
if(++r>n) break;
}
if(!(--cnt[a[l]])) sum--,vis[a[l]]=0;
l++;
}
printf("%d
",ans);
return 0;
}
❤bzoj1685：不懂系列！
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
int read(){
int x=0;char c=getchar();
while(!isdigit(c)) c=getchar();
while(isdigit(c)) x=x*10+c-'0',c=getchar();
return x;
}
const int nmax=25;
struct node{
int w,b;
bool operator<(const node&rhs)const{
return w<rhs.w;}
};
node ns[nmax];
int main(){
int n=read(),m=read();
rep(i,1,n) ns[i].w=read(),ns[i].b=read();
sort(ns+1,ns+n+1);
int ans=0,t,temp;
while(1){
temp=m;
dwn(i,n,1) {
t=min(temp/ns[i].w,ns[i].b);
temp-=t*ns[i].w;
ns[i].b-=t;
}
rep(i,1,n) if(ns[i].b>0&&temp>0){
ns[i].b--;
temp-=ns[i].w;
}//没办法了就扔多一个硬币给他，注意一个硬币就好了。
if(temp<=0) ans++;
else break;
}
printf("%d
",ans);
return 0;
}
bzoj1680：。。。
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
#define ll long long
int read(){
int x=0;char c=getchar();
while(!isdigit(c)) c=getchar();
while(isdigit(c)) x=x*10+c-'0',c=getchar();
return x;
}
const int nmax=10005;
const int inf=0x7f7f7f7f;
struct node{
int w,nd;
};
node ns[nmax];
void mins(int &a,int b){
if(a>b) a=b;
}
int main(){
int n=read(),s=read();
rep(i,1,n) ns[i].w=read(),ns[i].nd=read();
ll ans=0;int nmin=inf;
rep(i,1,n){
mins(nmin,ns[i].w);
ans+=ns[i].nd*nmin;
nmin+=s;
}
printf("%lld
",ans);
return 0;
}
bzoj1676：差分序列就好了。。。
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define clr(x,c) memset(x,c,sizeof(x))
#define dwn(i,s,t) for(int i=s;i>=t;i--)
int read(){
int x=0;char c=getchar();
while(!isdigit(c)) c=getchar();
while(isdigit(c)) x=x*10+c-'0',c=getchar();
return x;
}
const int nmax=2005;
int a[nmax];
void maxs(int &a,int b){
if(a<b) a=b;
}
int main(){
int n=read(),fa=read(),fb=read(),D=read(),lft=fa-fb,u,v,d,tmp=0,temp=0;
rep(i,1,n) {
u=read(),v=read();
a[v]++,a[u-1]--,maxs(tmp,v);
}
dwn(i,tmp,1){
temp+=a[i];
if(i<=D) lft-=temp;
if(!lft) {
printf("%d
",i);break;
}
}
return 0;
}
bzoj1675：搜索+hash判重！
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
const int nmax=100005;
const int mod=100007;
int head[mod],next[nmax],a[nmax],ans=0;
char map[6][6];bool vis[6][6];
int xx[5]={0,0,0,1,-1};
int yy[5]={0,1,-1,0,0};
void print(int x){
rep(i,1,5){
rep(j,1,5) printf("%d",x&1),x>>=1;printf("
");
}
}
void dfs(int x,int y,int hn,int jn,int pre){
if(map[x][y]=='H') hn++;else jn++;
if(hn>3) return ;
pre+=(1<<((x-1)*5+y-1));
int tmp=pre%mod;bool f=true;
for(int i=head[tmp];i;i=next[i]) if(a[i]==pre) f=false;
if(f) {
a[++a[0]]=pre;next[a[0]]=head[tmp];head[tmp]=a[0];
}
else return ;
if(hn+jn==7) ans++;
else{
vis[x][y]=1;
int tx,ty;
rep(i,1,5) rep(j,1,5) if(vis[i][j])
rep(k,1,4) {
tx=i+xx[k];ty=j+yy[k];
if(tx<1||ty<1||tx>5||ty>5||vis[tx][ty]) continue;
dfs(tx,ty,hn,jn,pre);
}
vis[x][y]=0;
}
}
int main(){
rep(i,1,5) scanf("%s",map[i]+1);
rep(i,1,5) rep(j,1,5) dfs(i,j,0,0,0);
printf("%d
",ans);
return 0;
}
bzoj1674：bfs。。。
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
#define qwq(x) for(edge *o=head[x];o;o=o->next)
int read(){
int x=0;char c=getchar();
while(!isdigit(c)) c=getchar();
while(isdigit(c)) x=x*10+c-'0',c=getchar();
return x;
}
const int nmax=1005;
const int maxn=50005;
const int inf=0x7f7f7f7f;
struct edge{
int to;edge *next;
};
edge edges[maxn],*pt=edges,*head[nmax];
void add(int u,int v){
pt->to=v;pt->next=head[u];head[u]=pt++;
}
int q[nmax],dist[nmax];
int main(){
int N=read(),T=read(),u,v;
rep(i,1,N) {
u=read(),v=read();
add(u,v);
}
int l=1,r=1;q[1]=1;
while(l<=r){
u=q[l];l++;
qwq(u) if(!dist[o->to]){
dist[o->to]=dist[u]+1;
q[++r]=o->to;
if(o->to==T) {
printf("%d
",dist[o->to]+1);
return 0;
}
}
}
printf("-1
");
return 0;
}
洛谷3936：挖坑。。。这道题我只写了三十分，mle了而且不mle的话也应该是50分的啊。。。我用hash判重+dp。
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
#define ll unsigned long long
const int nmax=1001;
ll pre[nmax][nmax],suf[nmax][nmax],a[nmax];
bool dp[nmax][nmax];
char s[nmax];
int main(){
scanf("%s",s+1);
int len=strlen(s+1);
rep(i,1,len) a[i]=s[i]-'a'+1;
rep(i,1,len) {
pre[i][i]=a[i];
rep(j,i+1,len) pre[i][j]=pre[i][j-1]*26+a[j];
}
dwn(i,len,1){
suf[i][i]=a[i];
dwn(j,i-1,1) suf[j][i]=suf[j+1][i]*26+a[j];
}
rep(i,1,len) dp[i][i]=1,dp[i][i+1]=a[i]==a[i+1]?1:0;
rep(i,2,len-1) rep(j,1,len-i) {
int tmp=i/2;
if(i%2==0) tmp--;
if(pre[j][j+tmp]==suf[j+i-tmp][j+i]) dp[j][j+i]=1;
}
int cur=1,ans=0;
while(cur<=len){
dwn(i,len,cur) if(dp[cur][i]) {
ans++;cur=i+1;
break;
}
if(cur>len) break;
}
printf("%d
",ans);
return 0;
}
洛谷3868：康托展开就好了
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(ll i=s;i<=t;i++)
#define dwn(i,s,t) for(ll i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
#define ll long long
ll read(){
ll x=0;char c=getchar();
while(!isdigit(c)) c=getchar();
while(isdigit(c)) x=x*10+c-'0',c=getchar();
return x;
}
const int nmax=105;
const ll mod=1000000007;
ll a[nmax];
bool vis[nmax];
int main(){
ll n=read(),u,v,d,tmp,temp;
a[0]=1;
rep(i,1,n) a[i]=(a[i-1]*i)%mod;
ll ans=1;
rep(i,1,n) {
u=read();v=0;
rep(j,1,u-1) if(!vis[j]) v++;
vis[u]=1;
ans=(ans+(v*a[n-i])%mod)%mod;
}
printf("%lld
",ans);
return 0;
}
bzoj2018：这
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(ll i=s;i<=t;i++)
#define dwn(i,s,t) for(ll i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
#define ll long long
ll read(){
ll x=0;char c=getchar();
while(!isdigit(c)) c=getchar();
while(isdigit(c)) x=x*10+c-'0',c=getchar();
return x;
}
const int nmax=1500005;
struct node{
ll w,u;
bool operator<(const node&rhs)const{
return u>rhs.u||u==rhs.u&&w<rhs.w;}
};
node ns[nmax];
int main(){
ll n=read(),a=read(),b=read(),c=read(),d=read(),e=read(),f=read(),g=read(),h=read(),m=read(),tmp,temp;
rep(i,0,n*3-1) {
tmp=(i*i)%d;
ns[i].w=((a*i%d)*(tmp*tmp%d)%d+b*tmp%d+c)%d;
tmp=(i*i)%h;
ns[i].u=((e*i%h)*(tmp*tmp%h)%h+(f*tmp%h)*i%h+g)%h;
}
sort(ns,ns+n*3);
ll ans=0;
rep(i,0,n-1) ans=(ans+ns[i].w)%m;
printf("%lld
",ans);
return 0;
}
❤bzoj3890：拓扑图上的dp。
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<bitset>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
#define qwq(x) for(edge *o=head[x];o;o=o->next)
int read(){
int x=0;char c=getchar();
while(!isdigit(c)) c=getchar();
while(isdigit(c)) x=x*10+c-'0',c=getchar();
return x;
}
const int nmax=105;
const int maxn=10005;
bitset<maxn>f[nmax],g[nmax];
struct edge{
int to,da,db;edge *next;
};
edge es[maxn],*pt=es,*head[nmax];
int in[nmax],q[nmax];
void add(int u,int v,int d,int w){
pt->to=v;pt->da=d;pt->db=w;pt->next=head[u];head[u]=pt++;
}
int main(){
int n=read(),m=read(),u,v,d,w,tmp,temp;
rep(i,1,m) u=read(),v=read(),d=read(),w=read(),add(u,v,d,w),in[v]++;
int l=1,r=0;
rep(i,1,n) if(!in[i]) q[++r]=i;
f[1][0]=g[1][0]=1;
while(l<=r){
u=q[l++];
qwq(u) {
f[o->to]|=f[u]<<o->da;
g[o->to]|=g[u]<<o->db;
if(!--in[o->to]) q[++r]=o->to;
}
}
rep(i,0,10000) if(f[n][i]&&g[n][i]){
printf("%d
",i);return 0;
}
printf("IMPOSSIBLE
");
return 0;
}
bzoj3480：先背包dp预处理后递推就好了
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
int read(){
int x=0;char c=getchar();
while(!isdigit(c)) c=getchar();
while(isdigit(c)) x=x*10+c-'0',c=getchar();
return x;
}
const int nmax=105;
const int maxn=100005;
const int inf=0x7f7f7f7f;
int a[nmax],dp[maxn];
void mins(int &a,int b){
if(a>b) a=b;
}
int main(){
int n=read(),m=read();
rep(i,1,m) a[i]=read();
clr(dp,0x7f);dp[0]=0;
rep(i,1,n) rep(j,a[i],100000) mins(dp[j],dp[j-a[i]]+1);
rep(i,1,n) a[i]=read();
int cur=1;
while(!a[cur]&&cur<=n) cur++;
int ans=dp[a[cur]];
rep(i,cur+1,n) ans+=dp[a[i]+1-a[i-1]];
printf("%d
",ans);
return 0;
}
bzoj3394：floyed一下就好了。。。
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
int read(){
int x=0;char c=getchar();
while(!isdigit(c)) c=getchar();
while(isdigit(c)) x=x*10+c-'0',c=getchar();
return x;
}
const int nmax=505;
const int inf=0x7f7f7f7f;
int dist[nmax][nmax],a[nmax];
void mins(int &a,int b){
if(a>b) a=b;
}
int main(){
int n=read(),p=read(),m=read(),u,v,d;
rep(i,1,p) a[i]=read();
clr(dist,0x7f);
rep(i,1,n) dist[i][i]=0;
rep(i,1,m) u=read(),v=read(),dist[u][v]=dist[v][u]=read();
rep(k,1,n) rep(i,1,n) rep(j,1,n)
if(dist[i][k]!=inf&&dist[k][j]!=inf) mins(dist[i][j],dist[i][k]+dist[k][j]);
int ans=inf,res,tmp;
rep(i,1,n){
tmp=0;
rep(j,1,p) tmp+=dist[i][a[j]];
if(tmp<ans) ans=tmp,res=i;
}
printf("%d
",res);
return 0;
}
bzoj3389：最少多少个区间就能够覆盖整个区间。贪心。
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
int read(){
int x=0;char c=getchar();
while(!isdigit(c)) c=getchar();
while(isdigit(c)) x=x*10+c-'0',c=getchar();
return x;
}
const int nmax=25005;
struct node{
int l,r;
bool operator<(const node&rhs)const{
return l<rhs.l||l==rhs.l&&r>rhs.r;}
};
node ns[nmax];
int main(){
int n=read(),T=read();
rep(i,1,n) ns[i].l=read(),ns[i].r=read();
sort(ns+1,ns+n+1);
if(ns[1].l!=1){
printf("-1
");return 0;
}
if(ns[1].r==T){
printf("1
");return 0;
}//没有特判这个然后就WA了。。。
int cur=1,t=ns[1].r,u,v,d,tmp,temp,ans=1;
while(cur<=n){
u=-1;v=t;
rep(i,cur+1,n) {
if(ns[i].l>t+1) break;
if(ns[i].r>v){
v=ns[i].r;u=i;
}
}
if(u!=-1) ans++,cur=u,t=v;
else break;
if(t==T) {
printf("%d
",ans);return 0;
}
}
printf("-1
");
return 0;
}
bzoj3399：贪心。
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
int read(){
int x=0;char c=getchar();
while(!isdigit(c)) c=getchar();
while(isdigit(c)) x=x*10+c-'0',c=getchar();
return x;
}
const int nmax=25005;
struct node{
int a[nmax],b[nmax];
};
node ns;
int main(){
int n=read(),X=read(),Y=read();
rep(i,1,n) ns.a[i]=read(),ns.b[i]=read();
sort(ns.a+1,ns.a+n+1);sort(ns.b+1,ns.b+n+1);
int ans=0;
rep(i,1,n){
if(ns.a[i]>ns.b[i]) ans+=Y*(ns.a[i]-ns.b[i]);
else ans+=X*(ns.b[i]-ns.a[i]);
}
printf("%d
",ans);
return 0;
}
bzoj3398：递推！不要想复杂了。
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
int read(){
int x=0;char c=getchar();
while(!isdigit(c)) c=getchar();
while(isdigit(c)) x=x*10+c-'0',c=getchar();
return x;
}
int dp[100001];
int main(){
int n=read(),m=read();
dp[0]=1;
rep(i,1,n) {
if(i>m+1) dp[i]=dp[i-1]+dp[i-m-1];else dp[i]=dp[i-1]+1;
if(dp[i]>=5000011) dp[i]-=5000011;
}
printf("%d
",dp[n]);
return 0;
}
bzoj3396：直接最大流。。。
#include<cstdio>
#include<cstring>
#include<queue>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
#define qwq(x) for(edge *o=head[x];o;o=o->next)
int read(){
int x=0;char c=getchar();
while(!isdigit(c)) c=getchar();
while(isdigit(c)) x=x*10+c-'0',c=getchar();
return x;
}
const int nmax=60;
const int maxn=1605;
const int inf=0x7f7f7f7f;
struct edge{
int to,cap;edge *next,*rev;
};
edge es[maxn],*pt=es,*head[nmax],*cur[nmax],*p[nmax];
int cnt[nmax],h[nmax],id[nmax];
void add(int u,int v,int d){
pt->to=v;pt->cap=d;pt->next=head[u];head[u]=pt++;
pt->to=u;pt->cap=0;pt->next=head[v];head[v]=pt++;
head[u]->rev=head[v];head[v]->rev=head[u];
}
int maxflow(int s,int t,int n){
clr(cnt,0);cnt[0]=n;clr(h,0);
int flow=0,a=inf,x=s;edge *e;
while(h[s]<n){
for(e=cur[x];e;e=e->next) if(e->cap>0&&h[e->to]+1==h[x]) break;
if(e){
a=min(a,e->cap);p[e->to]=cur[x]=e;x=e->to;
if(x==t){
while(x!=s) p[x]->cap-=a,p[x]->rev->cap+=a,x=p[x]->rev->to;
flow+=a,a=inf;
}
}else{
if(!--cnt[h[x]]) break;
h[x]=n;
for(e=head[x];e;e=e->next) if(e->cap>0&&h[x]>h[e->to]+1) cur[x]=e,h[x]=h[e->to]+1;
cnt[h[x]]++;
if(x!=s) x=p[x]->rev->to;
}
}
return flow;
}
int main(){
int n=0,m=read(),u,v,d,tmp,temp,t,s;
char S[5],T[5];
rep(i,1,m){
scanf("%s%s",S,T);d=read();
if(S[0]<='Z'&&S[0]>='A') {
tmp=S[0]-'A'+26+1;u=id[tmp];
if(!u) id[tmp]=u=++n;
if(S[0]=='A') s=u;
if(S[0]=='Z') t=u;
}else {
tmp=S[0]-'a'+1;u=id[tmp];
if(!u) id[tmp]=u=++n;
}
if(T[0]<='Z'&&T[0]>='A') {
tmp=T[0]-'A'+26+1;v=id[tmp];
if(!v) id[tmp]=v=++n;
if(T[0]=='A') s=v;
if(T[0]=='Z') t=v;
}else {
tmp=T[0]-'a'+1;v=id[tmp];
if(!v) id[tmp]=v=++n;
}
add(u,v,d);
}
printf("%d
",maxflow(s,t,n));
return 0;
}
bzoj3391：树形dp。。
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define clr(x,c) memset(x,c,sizeof(x))
#define qwq(x) for(edge *o=head[x];o;o=o->next)
int read(){
int x=0;char c=getchar();
while(!isdigit(c)) c=getchar();
while(isdigit(c)) x=x*10+c-'0',c=getchar();
return x;
}
const int nmax=10005;
int size[nmax],son[nmax],n;
struct edge{
int to;edge *next;
};
edge es[nmax<<1],*pt=es,*head[nmax];
void add(int u,int v){
pt->to=v;pt->next=head[u];head[u]=pt++;
pt->to=u;pt->next=head[v];head[v]=pt++;
}
void maxs(int &x,int y){
if(x<y) x=y;
}
void dfs(int x,int fa){
son[x]=1;
qwq(x) if(o->to!=fa){
dfs(o->to,x);
son[x]+=son[o->to];
maxs(size[x],son[o->to]);
}
maxs(size[x],n-son[x]);
}
int main(){
n=read();int u,v;
rep(i,1,n-1) u=read(),v=read(),add(u,v);
dfs(1,0);
int tmp=n/2,flag=1;
rep(i,1,n) if(size[i]<=tmp) flag=true,printf("%d
",i);
if(!flag) printf("NONE
");
return 0;
}
bzoj3390：最大生成树QAQ
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
int read(){
int x=0;char c=getchar();
while(!isdigit(c)) c=getchar();
while(isdigit(c)) x=x*10+c-'0',c=getchar();
return x;
}
const int nmax=1005;
const int maxn=20005;
struct edge{
int s,t,d;
bool operator<(const edge &rhs) const{
return d>rhs.d;}
};
edge es[maxn];
int fa[nmax];
int find(int x){
return fa[x]==x?x:fa[x]=find(fa[x]);
}
int main(){
int n=read(),m=read(),u,v,d;
rep(i,1,m) {
edge &o=es[i];
o.s=read(),o.t=read(),o.d=read();
}
sort(es+1,es+m+1);
rep(i,1,n) fa[i]=i;
int ta,tb,cnt=0,ans=0;
rep(i,1,m){
edge &o=es[i];
ta=find(o.s);tb=find(o.t);
if(ta!=tb) {
fa[ta]=tb;
ans+=o.d;
if(++cnt==n-1) break;
}
}
if(cnt!=n-1) printf("-1
");
else printf("%d
",ans);
return 0;
}
❤bzoj3315：dp+优化。利用特点减少掉一维！
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
int read(){
int x=0;char c=getchar();
while(!isdigit(c)) c=getchar();
while(isdigit(c)) x=x*10+c-'0',c=getchar();
return x;
}
const int nmax=1005;
struct node{
int x,w;
bool operator<(const node&rhs)const{
return x<rhs.x;}
};
node ns[nmax];
int dp[nmax][nmax],pre[nmax][nmax];
int as(int x){
return x<0?-x:x;
}
int main(){
int n=read();
rep(i,1,n) ns[i].x=read(),ns[i].w=read();
sort(ns+1,ns+n+1);
/*dp[1][1]=ns[1].w;
int ans=0;
rep(i,2,n) {
dp[i][i]=ns[i].w;
rep(j,1,i-1) {
rep(k,1,j) if(ns[i].x-ns[j].x>=ns[j].x-ns[k].x)
dp[i][j]=max(dp[i][j],dp[j][k]);
ans=max(ans,dp[i][j]+=ns[i].w);
}
}
reverse(ns+1,ns+n+1);
clr(dp,0);dp[1][1]=ns[1].w;
rep(i,2,n) {
dp[i][i]=ns[i].w;
rep(j,1,i-1) {
rep(k,1,j) if(as(ns[i].x-ns[j].x)>=as(ns[j].x-ns[k].x))
dp[i][j]=max(dp[i][j],dp[j][k]);
ans=max(ans,dp[i][j]+=ns[i].w);
}
}
printf("%d
",ans);
return 0;*/
dp[1][1]=pre[1][1]=ns[1].w;
int ans=0,p;
rep(i,2,n){
pre[i][i]=dp[i][i]=ns[i].w;p=i-1;
dwn(j,i-1,1){
while(p&&as(ns[i].x-ns[j].x)>=as(ns[j].x-ns[p].x)||p>j) p--;p++;
dp[i][j]=pre[p][j]+ns[i].w;
ans=max(ans,dp[i][j]);
pre[j][i]=max(dp[i][j],pre[j+1][i]);
}
}
reverse(ns+1,ns+n+1);
clr(dp,0);clr(pre,0);
dp[1][1]=pre[1][1]=ns[1].w;
rep(i,2,n){
pre[i][i]=dp[i][i]=ns[i].w;p=i-1;
dwn(j,i-1,1){
while(p&&as(ns[i].x-ns[j].x)>=as(ns[j].x-ns[p].x)||p>j) p--;p++;
dp[i][j]=pre[p][j]+ns[i].w;
ans=max(ans,dp[i][j]);
pre[j][i]=max(dp[i][j],pre[j+1][i]);
}
}
printf("%d
",ans);
return 0;
}
❤bzoj3314：单调队列就可以了。
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
int read(){
int x=0;char c=getchar();
while(!isdigit(c)) c=getchar();
while(isdigit(c)) x=x*10+c-'0',c=getchar();
return x;
}
const int nmax=50005;
struct node{
int x,h;
bool operator<(const node&rhs)const{
return x<rhs.x;}
};
node ns[nmax];
bool fa[nmax],fb[nmax];
int q[nmax];
int main(){
int n=read(),D=read();
rep(i,1,n) ns[i].x=read(),ns[i].h=read();
sort(ns+1,ns+n+1);
//rep(i,1,n) printf("%d %d
",ns[i].x,ns[i].h);
int l=1,r=0;
rep(i,1,n){
while(l<=r&&ns[q[r]].h<ns[i].h) r--;
q[++r]=i;
while(l<=r&&ns[q[l]].x<ns[i].x-D) l++;
if(ns[q[l]].h>=ns[i].h*2) fa[i]=1;
//rep(j,l,r) printf("%d ",q[j]);printf("
");
}
l=1,r=0;
int ans=0;
dwn(i,n,1){
while(l<=r&&ns[q[r]].h<ns[i].h) r--;
q[++r]=i;
while(l<=r&&ns[q[l]].x>ns[i].x+D) l++;
if(ns[q[l]].h>=ns[i].h*2) fb[i]=1;
if(fa[i]&&fb[i]) ans++;
//rep(j,l,r) printf("%d ",q[j]);printf("
");
}
printf("%d
",ans);
return 0;
}
bzoj3301：康托展开两种情况。
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
#define ll long long
ll read(){
ll x=0;char c=getchar();
while(!isdigit(c)) c=getchar();
while(isdigit(c)) x=x*10+c-'0',c=getchar();
return x;
}
const int nmax=25;
ll a[nmax],ans[nmax];bool vis[nmax];
char s[5];
int main(){
ll n=read(),m=read(),u,v,d,tmp,temp;
a[0]=1;
rep(i,1,n) a[i]=a[i-1]*i;
while(m--){
scanf("%s",s);
if(s[0]=='P'){
clr(vis,0);
u=read()-1;
rep(i,1,n){
v=u/a[n-i]+1;d=0;
rep(j,1,n){
if(!vis[j]) d++;
if(d==v) {
ans[i]=j;vis[j]=1;
break;
}
}
u%=a[n-i];
}
rep(i,1,n-1) printf("%lld ",ans[i]);printf("%lld
",ans[n]);
}else{
clr(vis,0);
rep(i,1,n) ans[i]=read();
u=1;
rep(i,1,n){
vis[ans[i]]=1;v=0;
rep(j,1,ans[i]-1) if(!vis[j]) v++;
u+=v*a[n-i];
}
printf("%lld
",u);
}
}
return 0;
}
bzoj3300：可以先预处理出每个括号各自的配对然后递归就可以了。
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define clr(x,c) memset(x,c,sizeof(x))
#define ll long long
int read(){
int x=0;char c=getchar();
while(!isdigit(c)) c=getchar();
while(isdigit(c)) x=x*10+c-'0',c=getchar();
return x;
}
const int nmax=100005;
const ll mod=12345678910;
int match[nmax],s[nmax];
ll dfs(int l,int r){
if(r==l+1) return 1;
if(match[l]==r) return (dfs(l+1,r-1)*2)%mod;
return (dfs(l,match[l])+dfs(match[l]+1,r))%mod;
}
int main(){
int n=read(),u,v,d,tmp=0,temp=0;
rep(i,1,n){
u=read();
if(!u) s[++tmp]=i;
else match[s[tmp--]]=i;
}
printf("%lld
",dfs(1,n));
return 0;
}
❤bzoj3297：string+dp。。。stl神器。。。
#include<cstdio>
#include<string>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
const int nmax=1005;
int len[nmax];
char s[nmax];
string ch[nmax],dp[nmax];
bool pd(int u,int x){
rep(i,0,len[x]-1)
if(s[u+i]!='?'&&s[u+i]!=ch[x][i]) return 0;
return 1;
}
int main(){
int n,m,u,v,d,tmp,temp;
scanf("%d%d",&n,&m);
scanf("%s",s+1);
rep(i,1,m) cin>>ch[i],len[i]=ch[i].size();
rep(i,1,n) rep(j,1,m){
u=i-len[j];
if(u<0) continue;
if(u&&dp[u]=="") continue;
if(pd(u+1,j))
if(dp[i]==""||dp[i]>dp[u]+ch[j]) dp[i]=dp[u]+ch[j];
}
cout<<dp[n]<<endl;
return 0;
}
bzoj3296：并查集。。。
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
int read(){
int x=0;char c=getchar();
while(!isdigit(c)) c=getchar();
while(isdigit(c)) x=x*10+c-'0',c=getchar();
return x;
}
const int nmax=10005;
const int maxn=30005;
int fa[nmax+maxn];
int find(int x){
return fa[x]==x?x:fa[x]=find(fa[x]);
}
int main(){
int n=read(),m=read(),u,v,ta,tb;
rep(i,1,n+m) fa[i]=i;
rep(i,1,n){
u=read();
rep(j,1,u) {
v=read();
ta=find(i);tb=find(n+v);
if(ta!=tb) fa[ta]=tb;
}
}
int ans=0;
rep(i,1,n) find(i);
sort(fa+1,fa+n+1);
rep(i,1,n) if(fa[i]!=fa[i-1]) ans++;
printf("%d
",ans-1);
return 0;
}
bzoj1681:最短路。。。
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
#define qwq(x) for(edge *o=head[x];o;o=o->next)
int read(){
int x=0;char c=getchar();
while(!isdigit(c)) c=getchar();
while(isdigit(c)) x=x*10+c-'0',c=getchar();
return x;
}
const int nmax=505;
const int maxn=2005;
const int inf=0x7f7f7f7f;
struct edge{
int to,dist;edge *next;
};
edge edges[maxn],*pt=edges,*head[nmax];
void add(int u,int v,int d){
pt->to=v;pt->dist=d;pt->next=head[u];head[u]=pt++;
pt->to=u;pt->dist=d;pt->next=head[v];head[v]=pt++;
}
struct node{
int x,dist;
node(int to,int dist):x(to),dist(dist){};
bool operator<(const node&rhs)const{
return dist>rhs.dist;}
};
priority_queue<node>q;
int dist[nmax],N,M,C,nd,tmp[nmax];
void dijkstra(){
clr(dist,0x7f);dist[1]=0;
q.push(node(1,0));
int tx,td;
while(!q.empty()){
node o=q.top();q.pop();
tx=o.x,td=o.dist;
if(dist[tx]!=td) continue;
qwq(tx) if(dist[o->to]>td+o->dist)
dist[o->to]=td+o->dist,q.push(node(o->to,dist[o->to]));
}
}
int main(){
N=read(),M=read(),C=read(),nd=read();
int u,v,d;
rep(i,1,M) u=read(),v=read(),d=read(),add(u,v,d);
dijkstra();
int ans=0;
rep(i,1,C){
u=read();
if(dist[u]<=nd) ans++,tmp[ans]=i;
}
printf("%d
",ans);
rep(i,1,ans) printf("%d
",tmp[i]);
return 0;
}
bzoj1668：递推。。。
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
const int nmax=105;
int map[nmax][nmax],dp[nmax][nmax];
int main(){
int n,m;scanf("%d%d",&n,&m);
rep(i,1,n) rep(j,1,m)
scanf("%d",&map[i][j]);
dp[1][1]=map[1][1];
rep(i,2,m) rep(j,1,i)
dp[j][i]=map[j][i]+max(dp[j][i-1],max(dp[j-1][i-1],dp[j+1][i-1]));
printf("%d
",dp[n][m]);
return 0;
}
bzoj1590：trie然后顺便维护一下子树末节点的个数就好了
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
int read(){
int x=0;char c=getchar();
while(!isdigit(c)) c=getchar();
while(isdigit(c)) x=x*10+c-'0',c=getchar();
return x;
}
const int nmax=500005;
int ch[nmax][2],size[nmax],F[nmax];
int main(){
int N=read(),M=read(),u,v,pt=0,t;
rep(i,1,N){
t=0;u=read();
rep(j,1,u) {
v=read();
if(!ch[t][v]) ch[t][v]=++pt;
t=ch[t][v];size[t]++;
}
F[t]++;//没有考虑重串WA了。
}
//rep(i,1,pt) printf("%d ",size[i]);
rep(i,1,M){
t=0;u=read();int ans=0;
bool f=true;
rep(j,1,u){
v=read();
if(!f) continue;
if(ch[t][v]) ans+=F[ch[t][v]],t=ch[t][v];
else f=false;
}
if(f) ans+=size[ch[t][0]]+size[ch[t][1]];
printf("%d
",ans);
}
return 0;
}
❤bzoj1597：dp+斜率优化。
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
#define ll long long
ll read(){
ll x=0;char c=getchar();
while(!isdigit(c)) c=getchar();
while(isdigit(c)) x=x*10+c-'0',c=getchar();
return x;
}
const int nmax=50005;
const ll inf=1e18;
ll dp[nmax];
struct node{
ll x,y;
bool operator<(const node&rhs)const {
return x>rhs.x||x==rhs.x&&y>rhs.y;}
};
node nodes[nmax];
ll x[nmax],y[nmax],q[nmax];
int main(){
int N=read(),u,v;
rep(i,1,N) nodes[i].x=read(),nodes[i].y=read();
sort(nodes+1,nodes+N+1);
int cnt=0;
rep(i,1,N) if(nodes[i].y>y[cnt]) {
x[++cnt]=nodes[i].x;y[cnt]=nodes[i].y;
}
int l=1,r=1;
rep(i,1,cnt){
while(l<r&&y[i]*(x[q[l]+1]-x[q[l+1]+1])>dp[q[l+1]]-dp[q[l]]) l++;
dp[i]=dp[q[l]]+x[q[l]+1]*y[i];
while(l<r&&(dp[q[r]]-dp[q[r-1]])*(x[q[r]+1]-x[i+1])>(dp[i]-dp[q[r]])*(x[q[r-1]+1]-x[q[r]+1])) r--;
q[++r]=i;
}
printf("%lld
",dp[cnt]);
return 0;
}
/*
5
100 1
15 15
15 13
20 5
1 100
*/
❤bzoj1690：分数规划！！！
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
#define qwq(x) for(edge *o=head[x];o;o=o->next)
int read(){
int x=0;char c=getchar();
while(!isdigit(c)) c=getchar();
while(isdigit(c)) x=x*10+c-'0',c=getchar();
return x;
}
const int nmax=1005;
const int maxn=5005;
const int inf=0x7f7f7f7f;
struct edge{
int to,dist;
double w;edge *next;
};
edge edges[maxn],*pt=edges,*head[nmax];
int a[nmax],tot[nmax],N,M;
double d[nmax];
bool inq[nmax];
void add(int u,int v,int d){
pt->to=v;pt->dist=d;pt->next=head[u];head[u]=pt++;
}
queue<int>q;
bool spfa(){
rep(i,1,N) d[i]=0,inq[i]=1,q.push(i),tot[i]=1;
while(!q.empty()){
int x=q.front();q.pop();inq[x]=0;
qwq(x) if(d[o->to]>d[x]+o->w){
d[o->to]=d[x]+o->w;
if(++tot[o->to]>=N) return 1;
if(!inq[o->to]) q.push(o->to),inq[o->to]=1;
}
}
return 0;
}
bool check(double L){
rep(i,1,N) qwq(i) o->w=L*o->dist-a[o->to];
return spfa();
}
int main(){
N=read(),M=read();int u,v,d;
rep(i,1,N) a[i]=read();
rep(i,1,M) u=read(),v=read(),d=read(),add(u,v,d);
double l=0,r=10000,mid;
while(r-l>0.001){
mid=(l+r)/2;
if(check(mid)) l=mid;
else r=mid;
}
printf("%.2lf
",l);
return 0;
}
bzoj1666：模拟。。。
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
int main(){
int n,ans=0;scanf("%d",&n);
while(n!=1){
ans++;
if(n%2==0) n>>=1;
else n=n*3+1;
}
printf("%d
",ans);
return 0;
}
bzoj1593：线段树经典
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define clr(x,c) memset(x,c,sizeof(x))
#define lson l,mid,x<<1
#define rson mid+1,r,x<<1|1
int read(){
int x=0;char c=getchar();
while(!isdigit(c)) c=getchar();
while(isdigit(c)) x=x*10+c-'0',c=getchar();
return x;
}
const int nmax=50005;
int lmax[nmax<<2],rmax[nmax<<2],omax[nmax<<2],col[nmax<<2];
void pushup(int x,int cnt){
lmax[x]=((lmax[x<<1]==(cnt-(cnt>>1)))?lmax[x<<1]+lmax[x<<1|1]:lmax[x<<1]);
rmax[x]=((rmax[x<<1|1]==(cnt>>1)?rmax[x<<1|1]+rmax[x<<1]:rmax[x<<1|1]));
omax[x]=max(rmax[x<<1]+lmax[x<<1|1],max(omax[x<<1],omax[x<<1|1]));
}
void build(int l,int r,int x){
col[x]=-1;lmax[x]=rmax[x]=omax[x]=r-l+1;
if(l==r) return ;
int mid=(l+r)>>1;
build(lson);build(rson);
}
void pushdown(int x,int cnt){
if(col[x]!=-1) {
col[x<<1]=col[x<<1|1]=col[x];
lmax[x<<1]=rmax[x<<1]=omax[x<<1]=col[x]?0:cnt-(cnt>>1);
lmax[x<<1|1]=rmax[x<<1|1]=omax[x<<1|1]=col[x]?0:(cnt>>1);
col[x]=-1;}
}
void update(int tl,int tr,int p,int l,int r,int x){
if(tl<=l&&tr>=r){
if(p) col[x]=p,lmax[x]=rmax[x]=omax[x]=0;
else col[x]=p,lmax[x]=rmax[x]=omax[x]=r-l+1;
}else{
pushdown(x,r-l+1);
int mid=(l+r)>>1;
if(tl<=mid) update(tl,tr,p,lson);
if(tr>mid) update(tl,tr,p,rson);
pushup(x,r-l+1);
}
}
int query(int cnt,int l,int r,int x){
if(l==r) return l;
pushdown(x,r-l+1);
int mid=(l+r)>>1;
if(omax[x<<1]>=cnt) return query(cnt,lson);
if(rmax[x<<1]+lmax[x<<1|1]>=cnt) return mid-rmax[x<<1]+1;
return query(cnt,rson);
}
int main(){
int N=read(),M=read(),u,v,d;
build(1,N,1);
rep(i,1,M){
u=read(),v=read();
if(u==1){
if(omax[1]<v) printf("0
");
else{
d=query(v,1,N,1);printf("%d
",d);
update(d,d+v-1,1,1,N,1);
}
}else{
d=read();
update(v,v+d-1,0,1,N,1);
}
}
return 0;
}
bzoj1592：花最少的代价把序列转换成不下降或不上升队列。dp+优化还是利用特点减掉了一维。
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
int read(){
int x=0;char c=getchar();
while(!isdigit(c)) c=getchar();
while(isdigit(c)) x=x*10+c-'0',c=getchar();
return x;
}
const int nmax=2005;
const int inf=0x7f7f7f7f;
int a[nmax],b[nmax];
int as(int x){
return x<0?-x:x;
}
int f[nmax][nmax],g[nmax][nmax];
int main(){
int n=read(),u,v;
rep(i,1,n) b[i]=a[i]=read();
sort(a+1,a+n+1);
rep(i,1,n) rep(j,1,n){
f[i][j]=g[i-1][j]+as(b[i]-a[j]);
if(j==1) g[i][j]=f[i][j];
else g[i][j]=min(f[i][j],g[i][j-1]);
}
int ans=g[n][n];
clr(g,0);clr(f,0);
reverse(b+1,b+n+1);
rep(i,1,n) rep(j,1,n){
f[i][j]=g[i-1][j]+as(b[i]-a[j]);
if(j==1) g[i][j]=f[i][j];
else g[i][j]=min(f[i][j],g[i][j-1]);
}
printf("%d
",min(g[n][n],ans));
return 0;
}
bzoj1596：树形dp/贪心的做法。。。辛辛苦苦写了那么长的dp结果那么短的贪心同样A了。。。
/*#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
#define qwq(x) for(edge *o=head[x];o;o=o->next)
int read(){
int x=0;char c=getchar();
while(!isdigit(c)) c=getchar();
while(isdigit(c)) x=x*10+c-'0',c=getchar();
return x;
}
const int nmax=10005;
const int maxn=20005;
const int inf=0x7f7f7f7f;
struct edge{
int to;edge *next;
};
edge edges[maxn],*pt=edges,*head[nmax];
int dp[nmax][3];
void add(int u,int v){
pt->to=v;pt->next=head[u];head[u]=pt++;
pt->to=u;pt->next=head[v];head[v]=pt++;
}
void dfs(int x,int fa){
dp[x][1]=1;dp[x][0]=inf;dp[x][2]=0;
int sum=0,flag=false,tmp=inf;
qwq(x) if(o->to!=fa){
int to=o->to;dfs(to,x);
dp[x][1]+=min(dp[to][0],min(dp[to][1],dp[to][2]));
dp[x][2]+=min(dp[to][0],dp[to][1]);
sum+=min(dp[to][1],dp[to][0]);
if(dp[to][1]<dp[to][0]) flag=true;
else tmp=min(tmp,dp[to][1]-dp[to][0]);
}
if(sum){
if(!flag) sum+=tmp;
dp[x][0]=sum;
}
}
int main(){
int N=read(),u,v;
rep(i,1,N-1) u=read(),v=read(),add(u,v);
dfs(1,0);
printf("%d
",min(dp[1][1],dp[1][0]));
return 0;
}*/
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
#define qwq(x) for(edge *o=head[x];o;o=o->next)
int read(){
int x=0;char c=getchar();
while(!isdigit(c)) c=getchar();
while(isdigit(c)) x=x*10+c-'0',c=getchar();
return x;
}
const int nmax=10005;
const int maxn=20005;
const int inf=0x7f7f7f7f;
struct edge{
int to;edge *next;
};
edge edges[maxn],*pt=edges,*head[nmax];
bool vis[nmax];
void add(int u,int v){
pt->to=v;pt->next=head[u];head[u]=pt++;
pt->to=u;pt->next=head[v];head[v]=pt++;
}
int ans=0;
void dfs(int x,int fa){
bool flag=false;
qwq(x) if(o->to!=fa){
dfs(o->to,x);
if(vis[o->to]) flag=true;
}
if(!flag&&!vis[x]&&!vis[fa]) vis[fa]=1,ans++;
}
int main(){
int N=read(),u,v;
rep(i,1,N-1) u=read(),v=read(),add(u,v);
dfs(1,0);
printf("%d
",ans);
return 0;
}
bzoj1652：跟1742相似，，都是套路啊
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
const int nmax=2005;
int a[nmax],f[nmax][nmax];
int main(){
int n;scanf("%d",&n);
rep(i,1,n) scanf("%d",&a[i]);
rep(i,1,n) {
for(int j=1;j+i-1<=n;j++){
f[j][j+i-1]=max(f[j+1][j+i-1]+(n-i+1)*a[j],f[j][j+i-2]+(n-i+1)*a[j+i-1]);
}
}
printf("%d
",f[1][n]);
return 0;
}
bzoj1684：模拟这
#include<cstdio>
double as(double x){
return x<0?-x:x;
}
int main(){
int N,D;
scanf("%d%d",&N,&D);
double ans=2.0,tmp=N*1.0/D;
int ta,tb,temp;
for(int i=1;i<=32767;i++){
temp=tmp*i;
if(as(tmp-temp*1.0/i)<ans&&i*N!=D*temp) ans=as(tmp-temp*1.0/i),ta=temp,tb=i;
if(as(tmp-(temp+1)*1.0/i)<ans&&i*N!=D*(temp+1)) ans=as(tmp-(temp+1)*1.0/i),ta=temp+1,tb=i;
}
printf("%d %d
",ta,tb);
return 0;
}
❤bzoj3892：dp。
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
int read(){
int x=0,f=1;char c=getchar();
while(!isdigit(c)){
if(c=='-') f=-1;c=getchar();
}
while(isdigit(c)) x=x*10+c-'0',c=getchar();
return x*f;
}
const int nmax=505;
const int inf=0x7f7f7f7f;
int dp[nmax][nmax],x[nmax],y[nmax];
int as(int x){
return x<0?-x:x;
}
int main(){
int N=read(),K=read();
rep(i,1,N) x[i]=read(),y[i]=read();
clr(dp,0x7f);dp[1][0]=0;
rep(i,2,N) {
for(int j=0;j<=i-2&&j<=K;j++){
rep(k,0,j)
dp[i][j]=min(dp[i][j],dp[i-k-1][j-k]+as(x[i]-x[i-k-1])+as(y[i]-y[i-k-1]));
}
}
printf("%d
",dp[N][min(K,N-2)]);
}