Bzoj1051 haoi2006 受欢迎的牛

比較基础的题目

首先边肯定是要反着加的

然后能够用tarjan一次求出来全部的scc

这个也就是缩点

然后看有几个点入度是0

这个点指得是强连通分量

假设入度为0 的点仅仅有一个就直接把这个点的点数输出

否则输0

没了

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <stack>
#define MAX 50009
#define rep(i,j,k) for(int i = j; i <= k; i++)

using namespace std;

int to[2 * MAX], next[2 * MAX], head[MAX];
int SccNum = 0, DfsClock = 0, dfn[MAX], low[MAX], num[MAX], In[MAX], in[MAX];
int tot = 0, scc[MAX], n, m;

stack <int> s;

inline void add (int x, int y)
{
	to[++tot] = y;
	next[tot] = head[x];
	head[x] = tot;
}

void tarjan (int x)
{
	dfn[x] = low[x] = ++DfsClock;
	in[x] = 1;
	s.push (x);
	for (int i = head[x]; i; i = next[i])
		if (!dfn[to[i]])
		{
			tarjan (to[i]);
			low[x] = min (low[x], low[to[i]]);
		}
		else
			if (in[to[i]])
				low[x] = min (low[x], dfn[to[i]]);

	if (low[x] == dfn[x])
	{
		SccNum++;
		while (1)
		{
			num[SccNum]++;
			int now = s.top();
			s.pop ();
			in[now] = 0;
			scc[now] = SccNum;
			if (now == x)
				break;
		}
	}
}

int main()
{
	scanf ("%d%d", &n, &m);
	rep (i, 1, m)
	{
		int a1, a2;
		scanf ("%d%d", &a1, &a2);
		add (a2, a1);
	}
	rep (i, 1, n)
		if (!dfn[i])
			tarjan (i);

	rep (i, 1, n)
		for (int j = head[i]; j; j = next[j])
			if (scc[i] != scc[to[j]])
				In[scc[to[j]]]++;

//	rep (i, 1, SccNum)
//		printf ("%d-- %d
", i, num[i]);

	int ans = 0, u = 0;
	rep (i, 1, SccNum)
		if (!In[i])
			ans = num[i], u++;
	if (u == 1)
		printf ("%d
", ans);
	else
		printf ("0
");
	return 0;
}