poj2987 求最大权闭合回路

建图差不多和以前做的差不多，就是最后询问这个闭合子图有多少个的时候，只要输出这个图的S集合，就是进行dfs能遍历到的点一定在S集合中，不能遍历到的点在T集合中

#include <iostream>
#include <algorithm>
#include <string.h>
#include <cstdio>
#include <vector>
#include <queue>
using namespace std;
const int maxn=5555;
const long long INF=1LL<<60;
typedef long long LL;
struct Dinic
{

    struct Edge{
      LL from,to,cap,flow;
      Edge(LL cfrom=0,LL cto=0,LL ccap=0,LL cflow=0)
      {
          from=cfrom;  to=cto; cap=ccap; flow=cflow;
      }
    };
    int n,m,s,t;
    vector<Edge>edges;
    vector<int>G[maxn];
    bool vis[maxn];
    int d[maxn];
    int cur[maxn];
    void init(int n)
    {
        m=0;
        this->n=n;
        for(int i=0; i<=n; i++)G[i].clear();
        edges.clear();
    }
    void addEdge(LL from,LL to, LL cap)
    {
        edges.push_back(Edge(from,to,cap,0) );
        edges.push_back(Edge(to,from,0,0));
        m+=2;
        G[from].push_back(m-2);
        G[to].push_back(m-1);
    }
    bool BFS()
    {
        memset(vis,0,sizeof(vis));
        queue<int>Q;
       Q.push(s);
      d[s]=0;
      vis[s]=1;
      while(!Q.empty())
        {
            int x=Q.front(); Q.pop();
            for(int i=0; i<G[x].size(); i++)
            {
                Edge &e = edges[G[x][i]];
                if(vis[e.to]==false&&e.cap>e.flow)
                {
                     vis[e.to]=1;
                     d[e.to]=d[x]+1;
                     Q.push(e.to);
                }
            }
        }
        return vis[t];
    }
    LL DFS(int x,LL a)
    {
        if(x==t||a==0)return a;
        LL flow=0,f;
        for(int &i=cur[x]; i<G[x].size(); i++)
        {
            Edge &e=edges[G[x][i]];
            LL  dd =min(a,e.cap-e.flow);
            if(d[x]+1==d[e.to]&&(f=DFS(e.to,dd))>0)
            {
                e.flow+=f;
                edges[G[x][i]^1].flow-=f;
                flow+=f;
                a-=f;
                if(a==0)break;
            }
        }
        return flow;
    }
    LL Maxflow(int s,int t)
    {
        this->s=s;this->t=t;
        LL flow=0;
        while(BFS())
        {
            memset(cur,0,sizeof(cur));
            flow+=DFS(s,INF);
        }
        return flow;
    }
    void find(int cur)
    {
        vis[cur]=true;
        for(int i=0; i<G[cur].size(); i++)
        {
            Edge &e=edges[G[cur][i]];
            if(vis[e.to]||e.cap<=e.flow)continue;
            find(e.to);
        }
    }
    int solve()
    {
        memset(vis,0,sizeof(vis));
        for(int i=0; i<G[s].size(); i++)
        {
            Edge &e=edges[G[s][i]];
            if(vis[e.to]||e.cap<=e.flow)continue;
            find(e.to);
        }
        int ans=0;
        for(int i=1; i<=n-2; i++)
            ans+=vis[i];
        return ans;
    }
}T;
int P[maxn];
int main()
{
     int n,m;
     while(scanf("%d%d",&n,&m)==2)
     {
         LL S1=0,S2=0;
         T.init(n+2);
         for(int i=1; i<=n; i++)
         {
             scanf("%d",&P[i]);
             if(P[i]>0){
                S1+=P[i];S2+=P[i];
                T.addEdge(n+1,i,P[i]);
             }else{
                S1+=-P[i];
                T.addEdge(i,n+2,-P[i]);
             }
         }
         for(int i=1; i<=m; i++)
         {
             int a,b;
             scanf("%d%d",&a,&b);
             T.addEdge(a,b,S1);
         }
         LL a1=T.Maxflow(n+1,n+2);
         LL a2=T.solve();
         printf("%I64d %I64d\n",a2,S2-a1);
     }
     return 0;
}