A：阅读理解。
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 1010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
int x=0,f=1;char c=getchar();
while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
return x*f;
}
int n,m,a[N],b[N];
signed main()
{
/*#ifndef ONLINE_JUDGE
freopen("a.in","r",stdin);
freopen("a.out","w",stdout);
const char LL[]="%I64d
";
#endif*/
n=read(),m=read();
for (int i=1;i<=n;i++) a[i]=read();
for (int i=1;i<=n;i++) b[i]=read();
if (a[1])
{
if (a[m]) {cout<<"YES";return 0;}
if (b[m])
{
for (int i=m+1;i<=n;i++)
if (a[i]&&b[i]) {cout<<"YES";return 0;}
}
}
cout<<"NO";
return 0;
//NOTICE LONG LONG!!!!!
}
B：即维护一个01序列有多少段1，瞎算算即可。
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 100010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
int x=0,f=1;char c=getchar();
while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
return x*f;
}
int n,m,k,ans;
ll a[N];
signed main()
{
/*#ifndef ONLINE_JUDGE
freopen("a.in","r",stdin);
freopen("a.out","w",stdout);
const char LL[]="%I64d
";
#endif*/
n=read(),m=read(),k=read();
for (int i=1;i<=n;i++) a[i]=read()-k;
for (int i=1;i<=n;i++)
if (a[i]>0&&a[i-1]<=0) ans++;
for (int i=1;i<=m;i++)
{
int op=read();
if (op==0)
{
printf("%d
",ans);
}
else
{
int x=read(),y=read();
if (a[x]<=0)
{
a[x]+=y;
if (a[x]>0) ans++,ans-=a[x-1]>0,ans-=a[x+1]>0;
}
}
}
return 0;
//NOTICE LONG LONG!!!!!
}
C：任取两人各一段连续+区间，容易发现起始位置之差在模gcd(ta,tb)意义下总是相同的。
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
int x=0,f=1;char c=getchar();
while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
return x*f;
}
int l1,r1,t1,l2,r2,t2;
ll cross(ll l1,ll r1,ll l2,ll r2)
{
if (l1>l2) swap(l1,l2),swap(r1,r2);
return max(min(r2,r1)-l2+1,0ll);
}
signed main()
{
/*#ifndef ONLINE_JUDGE
freopen("a.in","r",stdin);
freopen("a.out","w",stdout);
const char LL[]="%I64d
";
#endif*/
l1=read(),r1=read(),t1=read(),l2=read(),r2=read(),t2=read();
int x=gcd(t1,t2);
ll ans=cross(l1,r1,l2,r2);
int u=(l2-l1)/x;
ans=max(ans,cross(l1,r1,l2-1ll*u*x,r2-1ll*u*x));
u--;
ans=max(ans,cross(l1,r1,l2-1ll*u*x,r2-1ll*u*x));
u++,u++;
ans=max(ans,cross(l1,r1,l2-1ll*u*x,r2-1ll*u*x));
cout<<ans;
return 0;
//NOTICE LONG LONG!!!!!
}
D：一个比较坑的题。容易想到随便找一个需要改的字符串，对其最短修改串（即去掉前后缀极长不需修改的部分）进行修改，判一下是否对其它串也合法就做完了。但问题是这样并不一定最优，可能需要延长修改部分，达到避免对某些串修改的目的，如ac→bc a→a，不应选a→b而应选ac→bc。于是考虑先把所有不需要改的串去掉，需要改的串求出最短修改串，如果任意一对最短修改串不同即无解。否则由每个最短修改串尽量往两边延伸，得到合法的极长修改串，再对其进行检验即可。
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 3010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
int x=0,f=1;char c=getchar();
while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
return x*f;
}
int n,u,nxt[N],L[N],R[N],len[N];
char a[N][N],b[N][N],anss[N],anst[N];
bool flag[N];
signed main()
{
#ifndef ONLINE_JUDGE
freopen("a.in","r",stdin);
freopen("a.out","w",stdout);
const char LL[]="%I64d
";
#endif
n=read();
for (int i=1;i<=n;i++) scanf("%s",a[i]+1);
for (int i=1;i<=n;i++) scanf("%s",b[i]+1);
for (int i=1;i<=n;i++)
{
int m=strlen(a[i]+1);len[i]=m;
flag[i]=1;
for (int j=1;j<=m;j++)
if (a[i][j]!=b[i][j]) {flag[i]=0;break;}
if (!flag[i])
{
int t;
for (int k=1;k<=m;k++)
if (a[i][k]!=b[i][k])
{
t=m;
while (a[i][t]==b[i][t]) t--;
L[i]=k,R[i]=t;
if (!u) u=t-k+1;else if (t-k+1!=u) {cout<<"NO";return 0;}
break;
}
}
}
int l=0,r=0;
for (int i=1;i<=3001;i++)
{
bool f=0;
for (int j=1;j<=n;j++)
if (!flag[j]&&L[j]==i) {f=1;break;}
if (f) {l=i-1;break;}
char c;
for (int j=1;j<=n;j++)
if (!flag[j]) {c=a[j][L[j]-i];break;}
for (int j=1;j<=n;j++)
if (!flag[j]) if (c!=a[j][L[j]-i]) {f=1;break;}
if (f) {l=i-1;break;}
}
for (int i=1;i<=3001;i++)
{
bool f=0;
for (int j=1;j<=n;j++)
if (!flag[j]&&R[j]+i>len[j]) {f=1;break;}
if (f) {r=i-1;break;}
char c;
for (int j=1;j<=n;j++)
if (!flag[j]) {c=a[j][R[j]+i];break;}
for (int j=1;j<=n;j++)
if (!flag[j]) if (c!=a[j][R[j]+i]) {f=1;break;}
if (f) {r=i-1;break;}
}
for (int i=1;i<=n;i++)
if (!flag[i])
{
int v=L[i]-l;u+=l+r;
for (int j=1;j<=u;j++) anss[j]=a[i][v+j-1],anst[j]=b[i][v+j-1];
break;
}
nxt[0]=-1;
for (int i=1;i<=u;i++)
{
int j=nxt[i-1];
while (~j&&anss[j+1]!=anss[i]) j=nxt[j];
nxt[i]=j+1;
}
for (int i=1;i<=n;i++)
{
int cur=0,m=len[i];
for (int j=1;j<=m;j++)
{
while (~cur&&a[i][j]!=anss[cur+1]) cur=nxt[cur];
cur++;
if (cur==u)
{
for (int k=j-u+1;k<=j;k++) a[i][k]=anst[k-(j-u+1)+1];
break;
}
}
for (int j=1;j<=m;j++)
if (a[i][j]!=b[i][j]) {cout<<"NO";return 0;}
}
cout<<"YES"<<endl;
printf("%s
",anss+1);printf("%s
",anst+1);
return 0;
//NOTICE LONG LONG!!!!!
}
E：显然二分答案转换成01序列上的问题，然后问题就变为是否能从s个区间中选m个使其并区间的数字和>=k。显然可以合并掉有包含关系的区间后dp，具体地，设f[i][j]为前i个区间选j个时最多能有多少个1，考虑上个区间与当前区间是否有交，分两种情况转移，没有交直接取最大值，有交的情况用堆（或者单调队列？反正cf跑得快）维护。
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
#define ll long long
#define N 1510
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
int x=0,f=1;char c=getchar();
while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
return x*f;
}
int n,m,s,k,a[N],c[N],f[N][N],ans=-1;
bool flag[N];
struct data
{
int x,y;
bool operator <(const data&a) const
{
return x<a.x||x==a.x&&y<a.y;
}
bool operator ==(const data&a) const
{
return x==a.x&&y==a.y;
}
}b[N],d[N];
priority_queue<int> ins,del;
bool check(int u)
{
for (int i=1;i<=n;i++)
if (a[i]<=u) c[i]=1;else c[i]=0;
for (int i=1;i<=n;i++) c[i]+=c[i-1];
memset(f,0,sizeof(f));
for (int j=1;j<=s;j++)
{
int x=0,t=0;
while (!ins.empty()) ins.pop();
while (!del.empty()) del.pop();
for (int i=1;i<=m;i++)
{
while (x<i&&b[x+1].y<b[i].x) x++,del.push(f[x][j-1]-c[b[x].y]),t=max(t,f[x][j-1]);
while (!del.empty()&&ins.top()==del.top()) ins.pop(),del.pop();
f[i][j]=max(t+c[b[i].y]-c[b[i].x-1],c[b[i].y]+(ins.empty()?-1000000:ins.top()));
ins.push(f[i][j-1]-c[b[i].y]);
}
}
for (int i=1;i<=m;i++) if (f[i][s]>=k) return 1;
return 0;
}
signed main()
{
#ifndef ONLINE_JUDGE
freopen("a.in","r",stdin);
freopen("a.out","w",stdout);
const char LL[]="%I64d
";
#endif
n=read(),m=read(),s=read(),k=read();
for (int i=1;i<=n;i++) a[i]=read();
for (int i=1;i<=m;i++) b[i].x=read(),b[i].y=read();
sort(b+1,b+m+1);
m=unique(b+1,b+m+1)-b-1;
for (int i=1;i<=m;i++)
for (int j=1;j<=m;j++)
if (i!=j&&b[j].x<=b[i].x&&b[j].y>=b[i].y) {flag[i]=1;break;}
int t=0;
for (int i=1;i<=m;i++)
if (!flag[i]) d[++t]=b[i];
m=t;s=min(s,m);
for (int i=1;i<=m;i++) b[i]=d[i];
sort(b+1,b+m+1);
int l=1,r=1000000000;
while (l<=r)
{
int mid=l+r>>1;
if (check(mid)) ans=mid,r=mid-1;
else l=mid+1;
}
cout<<ans;
return 0;
//NOTICE LONG LONG!!!!!
}
F：冷静一下可以发现，由异或的消去性，这个题和树没有任何关系，求出每个点到根的路径的异或和后，就变为一堆数的问题。容易想到的做法是建一棵trie，二分答案，对每个数trie上跑一遍。时间复杂度O(nlog2v)，空间复杂度O(nlogv)，都炸掉了。
先考虑优化时间复杂度。trie还是要建的，改为按位考虑。对于当前位是否填1，记录所有需要考虑的对（即异或后与已确定位完全相同），算出下一位均填0和均填1的方案数（即异或的下一位填0），与k进行比较确定答案，如果填1就将k减掉该方案数。然后给每个对拓展下一位。注意到考虑第i位时，所有对都处于trie的第i层，并且每个点只会被考虑两次，所以时间复杂度O(nlogv)。
然后考虑将空间复杂度优化至线性。将所有数排序，就得到了所谓的隐式trie，因为排序后的一段区间就对应了trie上的一棵子树。跑同样的做法即可。
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
#define ll long long
#define N 1000010
ll read()
{
ll x=0,f=1;char c=getchar();
while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
return x*f;
}
int n,p[N],t,cnt,tot;
ll m,ans,a[N],b[N][2],c[N][2];
struct data{int to,nxt;ll len;
}edge[N];
void addedge(int x,int y,ll z){t++;edge[t].to=y,edge[t].nxt=p[x],edge[t].len=z,p[x]=t;}
void dfs(int k)
{
for (int i=p[k];i;i=edge[i].nxt)
{
a[edge[i].to]=a[k]^edge[i].len;
dfs(edge[i].to);
}
}
int jump(int x,int u,int k)
{
if (u==0) return ((a[x]>>k)&1)==0?x:-1;
for (int i=x;i<=n;i++)
{
if ((a[i]>>k+1)!=(a[x]>>k+1)) return -1;
if ((a[i]>>k)&1) return i;
}
return -1;
}
int query(int x,int u,int k)
{
x=jump(x,u,k);
if (x==-1) return 0;
for (int i=x+1;i<=n;i++)
if ((a[i]>>k)!=(a[x]>>k)) return i-x;
return n-x+1;
}
signed main()
{
#ifndef ONLINE_JUDGE
freopen("a.in","r",stdin);
freopen("a.out","w",stdout);
const char LL[]="%I64d
";
#endif
n=read(),m=read();
for (int i=2;i<=n;i++)
{
int x=read();ll y=read();
addedge(x,i,y);
}
dfs(1);sort(a+1,a+n+1);cnt++;b[cnt][0]=b[cnt][1]=1;
for (int k=62;~k;k--)
{
ll s=0;
for (int i=1;i<=cnt;i++)
s+=1ll*query(b[i][0],0,k)*query(b[i][1],0,k)+1ll*query(b[i][0],1,k)*query(b[i][1],1,k);
tot=0;ans<<=1;
if (s>=m)
{
for (int i=1;i<=cnt;i++)
{
if (query(b[i][0],0,k)&&query(b[i][1],0,k)) tot++,c[tot][0]=jump(b[i][0],0,k),c[tot][1]=jump(b[i][1],0,k);
if (query(b[i][0],1,k)&&query(b[i][1],1,k)) tot++,c[tot][0]=jump(b[i][0],1,k),c[tot][1]=jump(b[i][1],1,k);
}
}
else
{
for (int i=1;i<=cnt;i++)
{
if (query(b[i][0],0,k)&&query(b[i][1],1,k)) tot++,c[tot][0]=jump(b[i][0],0,k),c[tot][1]=jump(b[i][1],1,k);
if (query(b[i][0],1,k)&&query(b[i][1],0,k)) tot++,c[tot][0]=jump(b[i][0],1,k),c[tot][1]=jump(b[i][1],0,k);
}
m-=s;ans++;
}
cnt=tot;
for (int i=1;i<=cnt;i++) b[i][0]=c[i][0],b[i][1]=c[i][1];
}
cout<<ans;
return 0;
//NOTICE LONG LONG!!!!!
}
G：先考虑怎么求答案是否为0。平面图上下连通相当于对偶图左右不连通，考虑以凸多边形某个点标记Bob，那么每个圆会给该点带来一个形如圆角凸多边形的不合法区域，不合法区域有交则称两圆连通，如果这些不合法区域使得左右连通，说明答案>0。回到原题，做法类似，对圆之间的连通关系（及圆和边界的连通关系）建图后，即求最小割点数量，拆点跑最小割即可。
主要问题在于如何判断圆角凸多边形是否有交。真要带上圆角简直不可做，事实上可以转化为求两凸多边形的最近距离。对任一凸包取反后求闵可夫斯基和，看所得凸包中是否存在点到原点的距离<两圆半径之和即可。这得讨论一下原点是否在凸包中。然后我就因为这个写挂调了几天。自闭了。
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cassert>
using namespace std;
#define ll long long
#define int long long
#define N 410
#define S 0
#define T 401
#define in(i) (i*2-1)
#define out(i) (i*2)
#define vector point
#define p(x,y) (point){x,y}
#define O p(0,0)
int read()
{
int x=0,f=1;char c=getchar();
while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
return x*f;
}
int n,m,w,L,R,p[N],d[N],q[N],cur[N],t=-1,ans;
struct point
{
int x,y;
vector operator +(const vector&a) const
{
return (vector){x+a.x,y+a.y};
}
vector operator -(const vector&a) const
{
return (vector){x-a.x,y-a.y};
}
int operator *(const vector&a) const
{
return x*a.y-y*a.x;
}
int operator ^(const vector&a) const
{
return x*a.x+y*a.y;
}
}a[N],u[N<<1],v[N<<1],r[N<<1],Left;
struct circle{point o;int r;
}b[N];
struct data{int to,nxt,cap,flow;
}edge[N*N<<3];
void addedge(int x,int y,int z)
{
t++;edge[t].to=y,edge[t].nxt=p[x],edge[t].cap=z,p[x]=t;
t++;edge[t].to=x,edge[t].nxt=p[y],edge[t].cap=0,p[y]=t;
}
bool bfs()
{
memset(d,255,sizeof(d));d[S]=0;
int head=0,tail=1;q[1]=S;
do
{
int x=q[++head];
for (int i=p[x];~i;i=edge[i].nxt)
if (d[edge[i].to]==-1&&edge[i].flow<edge[i].cap)
{
d[edge[i].to]=d[x]+1;
q[++tail]=edge[i].to;
}
}while (head<tail);
return ~d[T];
}
int work(int k,int f)
{
if (k==T) return f;
int used=0;
for (int i=cur[k];~i;i=edge[i].nxt)
if (d[k]+1==d[edge[i].to])
{
int w=work(edge[i].to,min(f-used,edge[i].cap-edge[i].flow));
edge[i].flow+=w,edge[i^1].flow-=w;
if (edge[i].flow<edge[i].cap) cur[k]=i;
used+=w;if (used==f) return f;
}
if (used==0) d[k]=-1;
return used;
}
void dinic()
{
while (bfs())
{
memcpy(cur,p,sizeof(p));
ans+=work(S,N);
}
}
int len(vector a){return a^a;}
bool isin(point *r,int n)
{
for (int i=2;i<=n;i++)
if ((r[i]-r[i-1])*(O-r[i-1])<0) return 0;
return 1;
}
bool check(point a,point b,int r)
{
if (((b-a)^(O-a))>0&&((a-b)^(O-b))>0)
return ((b-a)*(O-a))*((b-a)*(O-a))<r*r*len(a-b);
else return 0;
}
int merge(point *u,point *v)
{
r[1]=u[1]+v[1];int tail=1;
int i=1,j=1;
while (i<=n||j<=n)
{
if (i>n) j++;
else if (j>n) i++;
else if ((u[i]+v[j+1]-r[tail])*(u[i+1]+v[j]-r[tail])>0) j++;
else i++;
point t=u[i]+v[j];
while (tail>1&&(t-r[tail-1])*(r[tail]-r[tail-1])>0) tail--;
r[++tail]=t;
}
return tail;
}
bool cross(circle x,circle y)
{
for (int i=1;i<=n;i++) u[i]=x.o-(a[i]-Left);
for (int i=1;i<=n;i++) v[i]=(a[i]-Left)-y.o;
int u_first=1;
for (int i=2;i<=n;i++) if (u[i].x<u[u_first].x||u[i].x==u[u_first].x&&u[i].y<u[u_first].y) u_first=i;
int v_first=1;
for (int i=2;i<=n;i++) if (v[i].x<v[v_first].x||v[i].x==v[v_first].x&&v[i].y<v[v_first].y) v_first=i;
for (int i=n+1;i<=2*n;i++) u[i]=u[i-n],v[i]=v[i-n];
int tail=merge(u+u_first-1,v+v_first-1);
if (isin(r,tail)) return 1;
for (int i=1;i<=tail;i++)
if (len(r[i])<(x.r+y.r)*(x.r+y.r)) return 1;
for (int i=1;i<tail;i++)
if (check(r[i],r[i+1],x.r+y.r)) return 1;
return 0;
}
signed main()
{
#ifndef ONLINE_JUDGE
freopen("g.in","r",stdin);
freopen("g.out","w",stdout);
const char LL[]="%I64d
";
#endif
n=read(),w=read();memset(p,255,sizeof(p));
Left=(point){N*N,N*N};L=N*N;R=-N*N;
for (int i=1;i<=n;i++)
{
a[i].x=read(),a[i].y=read();
L=min(L,a[i].x),R=max(R,a[i].x);
if (a[i].x<Left.x) Left=a[i];
}
m=read();
for (int i=1;i<=m;i++) b[i].o.x=read(),b[i].o.y=read(),b[i].r=read();
for (int i=1;i<=m;i++)
for (int j=i+1;j<=m;j++)
if (cross(b[i],b[j])) addedge(out(i),in(j),1),addedge(out(j),in(i),1);
for (int i=1;i<=m;i++)
{
if (b[i].o.x-(R-L)-b[i].r<0) addedge(S,in(i),1);
if (b[i].o.x+(R-L)+b[i].r>w) addedge(out(i),T,1);
}
for (int i=1;i<=m;i++) addedge(in(i),out(i),1);
dinic();
cout<<ans;
return 0;
//NOTICE LONG LONG!!!!!
}