POJ2186 POPULAR COW

链接：http://poj.org/problem?id=2186

题意：给你N个点，然后在给你N条有向边，然后让你找出这样的点S,S满足条件图上任意一点都能到达S。

要想满足任意一点都能到达，首先满足图连通，然后满足将图缩点后形成一个棵树，树的特征是可以有且只有一个点的出度为0。

然后代码如下：

 #include <iostream>
 #include <vector>
 #include <queue>
 #include <string.h>
 #include <stdlib.h>
 #include <stdio.h>
 #include <stack>

 using namespace std;
 const int maxn = ;

 struct edge
 {
     int u,v,val;
 };
 int dfn[],low[],belong[],inst[];
 vector<edge> edges;
 vector<int>g[maxn];
 stack<int>st;
 int bcnt,cnt,time;

 void init(int n)
 {
     int i;
     for(i =;i <= n;i++)
     g[i].clear();

     edges.clear();
     time = ;bcnt = cnt = ;
     return ;
 }
 void addedge(int u,int v,int val)
 {
     edges.push_back((edge){u,v,});
     g[u].push_back(cnt);
     cnt++;

     return ;
 }
 void tarjan(int i)
 {
     int j;
     dfn[i] = low[i] = ++time;
     inst[i] = ;
     st.push(i);

     for(j = ;j < g[i].size();j++)
     {
         edge e;
         e = edges[g[i][j]];
         int v;
         v = e.v;
         if(!dfn[v])
         {
             tarjan(v);
             low[i] = min(low[i],low[v]);
         }
         else if(inst[v] && dfn[v] < low[i])
         low[i] = dfn[v];
     }
     if(dfn[i] == low[i])
     {
         //cout<<i<<"****"<<endl;
         bcnt++;
         do
         {
             j = st.top();
             st.pop();
             inst[j] = ;
             belong[j] = bcnt;
             //printf("%d *",j);
         }
         while(j != i);
         //puts("");
     }

 }
 void tarjans(int n)
 {
     int i;
     bcnt = time = ;
     while(!st.empty())st.pop();
     memset(dfn,,sizeof(dfn));

     memset(inst,,sizeof(inst));
     memset(belong,,sizeof(belong));
     for(i = ;i <= n;i++)
     if(!dfn[i])tarjan(i);
 }
 int main()
 {
     int n,m;
     //freopen("in.txt","r",stdin);
     while(~scanf("%d %d",&n,&m))
     {
         int a,b,i;
         init(n);
         for(i = ;i < m;i++)
         {
             scanf("%d %d",&a,&b);
             addedge(a,b,);

         }
         tarjans(n);
         //cout<<bcnt<<endl;
         int leap = ,flag = ;
         int bcnt0,j;
         int outbc[];
         memset(outbc,,sizeof(outbc));
         for(i = ;i <= n;i++)
         {
             for(j = ;j < g[i].size();j++)
             {
                 int v;
                 v = edges[g[i][j]].v;
                 if(belong[i] != belong[v]){
                 if(outbc[belong[i]] == )
                 leap++;
                 outbc[belong[i]] = ;
                 }
             }
         }
         if(leap == bcnt-)
         {
             int ans;
             ans = ;
             for(i = ;i < bcnt+;i++)
             {
                 if(outbc[i] == )
                 break;
             }
             for(j = ;j <= n;j++)
             if(belong[j] == i)
                 ans++;

             printf("%d\n",ans);
         }
         else
         puts("");

     }
     return ;
 }