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P1345 奶牛的电信

题目略，就是求最小割（容量和最小的边集，使得图不联通，常用S-T割)

那么最小割=最大流

这里要求点权和最小，可以通过拆点转化为边权

#include<iostream>
#include<algorithm>
#include<cstring>
#include<queue>
#define IOS std::ios::sync_with_stdio(0)
using namespace std;
const int  N=1e4,M=5e5;
 #define inf 1e9 
 int n,m;
  
 int all=1,hd[N],go[M],w[M],nxt[M];
    
 int S,T;
 int dis[M],ans=0,now[M];
    
 void add_(int x,int y,int z){
     nxt[++all]=hd[x]; hd[x]=all; go[all]=y;
     w[all]=z;
     swap(x,y);
     nxt[++all]=hd[x]; hd[x]=all; go[all]=y;
     w[all]=0;
 }
  bool bfs(){
     for(int i=0;i<M;i++)dis[i]=inf;
     queue<int> q;
     q.push(S);
     now[S]=hd[S];
     dis[S]=0;
        
     while(q.empty()==0){
         int x=q.front();
         q.pop();
         for(int i=hd[x];i;i=nxt[i]){
             int y=go[i];
             if(w[i]>0&&dis[y]==inf){
                 dis[y]=dis[x]+1;
                 now[y]=hd[y];
                 q.push(y);
                 if(y==T) return 1;
             }
         }
     }
     return 0;
 }
 int dfs(int x,int sum){
     if(x==T) return sum;
     int k,res=0;
        
     for(int i=now[x];i&& sum ;i=nxt[i]){
         now[x]=i;
         int y=go[i];
         if(w[i]>0&&(dis[y]==dis[x]+1)){
             k=dfs(y,min(sum,w[i]));
             if(k==0) dis[y]=inf;
             w[i]-=k;
             w[i^1]+=k;
             res+=k;
             sum-=k;
         }
     }
     return res;
 }
  
 void dinic(){
    int ans =0;
    while(bfs()) ans+=dfs(S,inf);   
    cout<<ans<<endl;
 }
  
 signed main(){
    int i,x,y;
    cin>>n>>m>>S>>T; 
    S+=n;
    for(i=1;i<=n;i++) add_(i,i+n,1);
         
    for(i=1;i<=m;i++){
         cin>>x>>y,add_(x+n,y,inf); add_(y+n,x,inf);
    }    
    dinic();
 }