POJ 2186 Popular Cows (Tarjan 有向图，强联通分量)

 求所有点都能够到达的点。

也就是强联通缩点之后，找出度为 0  的点。

如果出度为 0 的点之后一个，那这个点所在的圈的所有点的数量就是答案。

如果，出度为 0 的点不止一个，那么就无解 输出 0 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <stack>
#include <algorithm>
#include <vector>
#define mem(x,v) memset(x,v,sizeof(x)) 
#define rep(i,a,b)  for (int i = a; i < b; i++)
#define per(i,a,b)  for (int i = a; i > b; i--)
using namespace std;
typedef long long LL;
const double EPS = 1e-10;
const int INF = 0x3f3f3f3f;
const int N = 1e4+10;
const int M = 1e5+10;
stack<int>q;
int cnt = -1;
int Head[N],Next[M],p[M],low[N],scc[N];
bool vis[N];
int n,m,times,cont,dfn[N];
void Add_edge(int u, int v){
	++cnt;
	Next[cnt] = Head[u];
	Head[u] = cnt;
	p[cnt] = v;
	return;
}
void init(){
	int x,y;
	mem(Head,-1);
	scanf("%d%d",&n,&m);
	rep(i,0,m){
		scanf("%d%d",&x,&y);
		Add_edge(x,y);
	}
	return;
}
void Tarjan(int u){
	low[u] = dfn[u] = ++times;
	q.push(u);
	for (int i = Head[u]; i != -1; i = Next[i]){
		int v = p[i];
		if (!dfn[v]){
			Tarjan(v);
			low[u] = min(low[u],low[v]);
		} else 
		if (!scc[v]) low[u] = min(low[u],dfn[v]);
	}
	if (low[u] == dfn[u]){
		cont++;
		while(true){
			int v = q.top(); q.pop();
			scc[v] = cont;
			if (v == u) break;
		}
	}
	return;
}
int main(){
	init();
	rep(i,0,n+1){
		low[i] = scc[i] = dfn[i] = 0;
	}
	while(!q.empty()) q.pop();
	times = 0;
	cont = 0;
	rep(i,1,n+1){
		if (dfn[i] == 0) Tarjan(i);
	}
	vector<int>f[N];
	rep(i,1,cont+1)
	f[i].clear();
	rep(i,1,n+1){
		for (int j = Head[i]; j != -1; j = Next[j]){
			if (scc[i] != scc[p[j]]) f[scc[i]].push_back(scc[p[j]]);
		}
	}


	int t,Ans = 0,y = 0;
	rep(i,1,cont+1){
		if (f[i].size() == 0) {
			t = i;
			y++;
		}
	}
	if (y > 1) {
		printf("0\n");
		return 0;
	}
	rep(i,1,n+1)
	if (scc[i] == t) Ans++;
	printf("%d\n",Ans);
	return 0;
}