洛谷P1345 [USACO5.4]奶牛的电信Telecowmunication

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一眼就能看出是最小割啊。。
拆完点之后跑一边Dinic就行
拆出的点之前连边权为1的边
题目中给出的连边连INF的边
关键是代码怎么写啊QWQ
想了半天没想出来，
看了一下学姐的发现了一种很巧妙的方法

#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
const int MAXN=801,INF=5*1e8+10;
inline char nc()
{
    static char buf[MAXN],*p1=buf,*p2=buf;
    return p1==p2&&(p2=(p1=buf)+fread(buf,1,MAXN,stdin),p1==p2)?EOF:*p1++;
}
inline int read()
{
    char c=nc();int x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=nc();}
    while(c>='0'&&c<='9'){x=x*10+c-'0';c=nc();}
    return x*f;
}
int S=0,T=3001;
struct node
{
    int u,v,flow,nxt;
}edge[MAXN*20];
int head[MAXN],cur[MAXN],num=0;
inline void add_edge(int x,int y,int z)
{
    edge[num].u=x;
    edge[num].v=y;
    edge[num].flow=z;
    edge[num].nxt=head[x];
    head[x]=num++;
}
inline void AddEdge(int x,int y,int z) 
{
    add_edge(x,y,z);
    add_edge(y,x,0);
}int deep[MAXN];
inline bool BFS()
{
    memset(deep,0,sizeof(deep));
    deep[S]=1;
    queue<int>q;
    q.push(S);
    while(q.size()!=0)
    {
        int p=q.front();
        q.pop();
        for(int i=head[p];i!=-1;i=edge[i].nxt)
            if(!deep[edge[i].v]&&edge[i].flow)
            {
                deep[edge[i].v]=deep[p]+1;q.push(edge[i].v);
                if(edge[i].v==T) return 1;
            }
    }
    return deep[T];
}
int DFS(int now,int nowflow)
{
    if(now==T||nowflow<=0)    return nowflow;
    int totflow=0;
    for(int &i=cur[now];i!=-1;i=edge[i].nxt) 
    {
        if(deep[edge[i].v]==deep[now]+1&&edge[i].flow)
        {
            int canflow=DFS(edge[i].v,min(nowflow,edge[i].flow));
            edge[i].flow-=canflow;edge[i^1].flow+=canflow;
            totflow+=canflow;
            nowflow-=canflow;
            if(nowflow<=0) break;
        }
    }
    return totflow;
}
int Dinic()
{
    int ans=0;
    while(BFS())
    {
        memcpy(cur,head,sizeof(head)); 
        ans+=DFS(S,INF);
    }
    return ans;
}
int main()
{
    #ifdef WIN32
    freopen("a.in","r",stdin);
    #else
    #endif
    memset(head,-1,sizeof(head));
    int N=read(),M=read(),A=read(),B=read();
       for(int i=1;i<=N;i++) AddEdge(i+N,i,1);
    for(int i=1;i<=M;i++)
    {
        int x=read(),y=read();
        AddEdge(x,y+N,INF);
        AddEdge(y,x+N,INF);
    }
    S=A;T=B+N;
    printf("%d",Dinic());
    return  0;
}