【luogu P2341 [HAOI2006]受欢迎的牛】 题解

题解报告：https://www.luogu.org/problemnew/show/P2341
我们把图中的强连通分量缩点，然后只有出度为0的牛是受欢迎的，这样如果出度为0的牛只有一个，说明受所有牛欢迎。否则出度为0只是受一些牛欢迎。

#include <stack>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn = 100000 + 10;
struct edge{
	int next, to, from, len;
}e[maxn<<2];
int head[maxn], cnt;
int n, m, color[maxn], tong[maxn], du[maxn], num, tim, ans, tmp;
int dfn[maxn], low[maxn];
bool vis[maxn];
stack<int> s;
void add(int u, int v)
{
	e[++cnt].from = u;
	e[cnt].to = v;
	e[cnt].next = head[u];
	head[u] = cnt;
}
void tarjan(int x)
{
	dfn[x] = low[x] = ++tim;
	s.push(x); vis[x] = 1;
	for(int i = head[x]; i != -1; i = e[i].next)
	{
		int v = e[i].to;
		if(!dfn[v])
		{
			tarjan(v);
			low[x] = min(low[x], low[v]);
		}
		else if(vis[v])
		{
			low[x] = min(low[x], low[v]);
		}
	}
	if(dfn[x] == low[x])
	{
		color[x] = ++num;
		vis[x] = 0;
		while(s.top() != x)
		{
			color[s.top()] = num;
			vis[s.top()] = 0;
			s.pop(); 
		}
		s.pop();
	}
}
int main()
{
	memset(head, -1, sizeof(head));
	scanf("%d%d",&n,&m);
	for(int i = 1; i <= m; i++)
	{
		int u, v;
		scanf("%d%d",&u,&v);
		add(u,v);
	}
	for(int i = 1; i <= n; i++)
		if(!dfn[i]) tarjan(i);
	for(int i = 1; i <= n; i++)
	{
		for(int j = head[i]; j != -1; j = e[j].next)
		{
			int v = e[j].to;
			if(color[v] != color[i])
			{
				du[color[i]]++;
			}
		}
		tong[color[i]]++;
	}
	for(int i = 1; i <= num; i++)
	{
		if(du[i] == 0)
		{
			tmp++;
			ans = tong[i];
		}
	}
	if(tmp == 0) 
	{
		printf("0");
		return 0;
	}
	if(tmp == 1)
	{
		printf("%d",ans);
		return 0;
	}
	if(tmp > 1)
	{
		printf("0");
		return 0;
	}
}