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=;

const long long INF=1LL<<;

typedef long long LL;

struct Dinic

{

    struct Edge{

      LL from,to,cap,flow;

      Edge(LL cfrom=,LL cto=,LL ccap=,LL cflow=)

      {

          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=;

        this->n=n;

        for(int i=; i<=n; i++)G[i].clear();

        edges.clear();

    }

    void addEdge(LL from,LL to, LL cap)

    {

        edges.push_back(Edge(from,to,cap,) );

        edges.push_back(Edge(to,from,,));

        m+=;

        G[from].push_back(m-);

        G[to].push_back(m-);

    }

    bool BFS()

    {

        memset(vis,,sizeof(vis));

        queue<int>Q;

       Q.push(s);

      d[s]=;

      vis[s]=;

      while(!Q.empty())

        {

            int x=Q.front(); Q.pop();

            for(int i=; i<G[x].size(); i++)

            {

                Edge &e = edges[G[x][i]];

                if(vis[e.to]==false&&e.cap>e.flow)

                {

                     vis[e.to]=;

                     d[e.to]=d[x]+;

                     Q.push(e.to);

                }

            }

        }

        return vis[t];

    }

    LL DFS(int x,LL a)

    {

        if(x==t||a==)return a;

        LL flow=,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]+==d[e.to]&&(f=DFS(e.to,dd))>)

            {

                e.flow+=f;

                edges[G[x][i]^].flow-=f;

                flow+=f;

                a-=f;

                if(a==)break;

            }

        }

        return flow;

    }

    LL Maxflow(int s,int t)

    {

        this->s=s;this->t=t;

        LL flow=;

        while(BFS())

        {

            memset(cur,,sizeof(cur));

            flow+=DFS(s,INF);

        }

        return flow;

    }

    void find(int cur)

    {

        vis[cur]=true;

        for(int i=; 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,,sizeof(vis));

        for(int i=; 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=;

        for(int i=; i<=n-; i++)

            ans+=vis[i];

        return ans;

    }

}T;

int P[maxn];

int main()

{

     int n,m;

     while(scanf("%d%d",&n,&m)==)

     {

         LL S1=,S2=;

         T.init(n+);

         for(int i=; i<=n; i++)

         {

             scanf("%d",&P[i]);

             if(P[i]>){

                S1+=P[i];S2+=P[i];

                T.addEdge(n+,i,P[i]);

             }else{

                S1+=-P[i];

                T.addEdge(i,n+,-P[i]);

             }

         }

         for(int i=; i<=m; i++)

         {

             int a,b;

             scanf("%d%d",&a,&b);

             T.addEdge(a,b,S1);

         }

         LL a1=T.Maxflow(n+,n+);

         LL a2=T.solve();

         printf("%I64d %I64d\n",a2,S2-a1);

     }

     return ;

}