**********强连通分量************

修改low值与缩点 ----------------->见这里

POJ 2186 Popular Cows

Description

Every cow's dream is to become the most popular cow in the herd. In a herd of N (1 <= N <= 10,000) cows, you are given up to M (1 <= M <= 50,000) ordered pairs of the form (A, B) that tell you that cow A thinks that cow B is popular. Since popularity is transitive, if A thinks B is popular and B thinks C is popular, then A will also think that C is
popular, even if this is not explicitly specified by an ordered pair in the input. Your task is to compute the number of cows that are considered popular by every other cow.

Input

* Line 1: Two space-separated integers, N and M

* Lines 2..1+M: Two space-separated numbers A and B, meaning that A thinks B is popular.

Output

* Line 1: A single integer that is the number of cows who are considered popular by every other cow.

Sample Input

Sample Output

Hint

Cow 3 is the only cow of high popularity.

Source

USACO 2003 Fall

缩点就是在求出图的所有的强连通分量之后,把强连通分量都看成一个点,这样形成的一个先用tarjan求出每个强连通分量，再缩点，统计每个点的出度，如果有且只有1个出度为0的点，就输出这个点包含的节点数，否则输出0.

证明：

1.如果有强连通分量被孤立（即和其他强连通分量无边相连），那么答案一定是0，此时由于缩点后是一个DAG图，出度为0的点的个数一定大于1.

2.如果没有点被孤立，当出度为0的点多于1个时，由DAG图的性质可得，一定不存在一个点能从其他所有点到达。只有当出度为0的点的个数等于1时，这个出度为0的点才能被其他所有点到达。

还是应多做题目，加深理解过程。

#include<stdio.h>
#include<stdlib.h>
#include<iostream>
#include<algorithm>
#include<vector>
#include<string.h>
using namespace std;
const int maxn=11000;
vector<int>mp[maxn];
int vis[maxn];
int dfn[maxn],low[maxn],S[maxn],color[maxn],sum[maxn],out0[maxn];
int n,m,cnt,tt,sig;
void Tarjan(int u)
{
	vis[u]=1;
	dfn[u]=low[u]=++cnt;
	S[++tt]=u;
	for(int i=0;i<mp[u].size();i++)
	{
		int v=mp[u][i];
		if(vis[v]==0)
		{
			Tarjan(v);
		}
		if(vis[v]!=0)
		{
			low[u]=min(low[u],low[v]);
		}
	}
	if(dfn[u]==low[u])
	{
		int len=tt;
		sig++;
	   do
	   {
	   	color[S[tt]]=sig;
	   }while(S[tt--]!=u);
	    sum[sig]=len-tt;
	}
}
void Solve()
{
	tt=-1,sig=0,cnt=0;
	for(int i=1;i<=n;i++)
	{
		if(vis[i]==0)
		{
			Tarjan(i);
		}
	}
	for(int i=1;i<=n;i++)
	{
		for(int j=0;j<mp[i].size();j++)
		{
			int v=mp[i][j];
			if(color[i]!=color[v])
			{
				out0[color[i]]++;
			}
		}
	}
	int p=0;
	for(int i=0;i<=sig;i++)
	{
		if(!out0[i])
		{
			if(p>0)
			{
				puts("0");
				return ;
			}
			p=sum[i];
		}
	}
	printf("%d\n",p);
}
void init()
{
    memset(vis,0,sizeof(vis));
    memset(dfn,0,sizeof(dfn));
    memset(low,0,sizeof(low));
    memset(mp,0,sizeof(mp));
    memset(out0,0,sizeof(out0));
}
int main()
{
	scanf("%d%d",&n,&m);
    init();
    int u,v;
    for(int i=0;i<m;i++)
    {
    	scanf("%d%d",&u,&v);
    	mp[u].push_back(v);
	}
	Solve();
	return 0;
}