【图论】Popular Cows

[POJ2186]Popular Cows

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 34752		Accepted: 14155

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

3 3
1 2
2 1
2 3

Sample Output

1

Hint

Cow 3 is the only cow of high popularity. 

Source

USACO 2003 Fall

 

题目大意：给定一个有向图，求出有多少点满足所有点可以间接或直接地到它。

试题分析：Tarjan缩点，然后求出哪个点出度为0就好了，输出其大小。(因为缩点后是DAG)

　　　　　如果有>1个出度为0那么就肯定没有了。

 

代码：

#include<iostream>
#include<cstring>
#include<cstdio>
#include<vector>
#include<queue>
#include<stack>
#include<algorithm>
using namespace std;

inline int read(){
	int x=0,f=1;char c=getchar();
	for(;!isdigit(c);c=getchar()) if(c=='-') f=-1;
	for(;isdigit(c);c=getchar()) x=x*10+c-'0';
	return x*f;
}
const int MAXN=500001;
const int INF=999999;
int N,M;
int dfn[MAXN],low[MAXN];
int que[MAXN]; bool vis[MAXN];
vector<int> vec[MAXN];
int outdu[MAXN];
int tar[MAXN];
int tot,tmp,Col;
int size[MAXN];

void Tarjan(int x){
	++tot; dfn[x]=low[x]=tot;
	vis[x]=true; que[++tmp]=x;
	for(int i=0;i<vec[x].size();i++){
		int to=vec[x][i];
		if(!dfn[to]){
			Tarjan(to);
			low[x]=min(low[x],low[to]);
		}
		else if(vis[to]) low[x]=min(dfn[to],low[x]);
	}
	if(dfn[x]==low[x]){
		++Col; tar[x]=Col;
		vis[x]=false;
		while(que[tmp]!=x){
			int k=que[tmp];
			tar[k]=Col; vis[k]=false;
			tmp--;
		}
		tmp--;
	}
}
int ans,anst;
int main(){
    N=read(),M=read();
	for(int i=1;i<=M;i++){
		int u=read(),v=read();
		vec[u].push_back(v);
	}
	for(int i=1;i<=N;i++) if(!dfn[i]) Tarjan(i);
	for(int i=1;i<=N;i++){
		int col=tar[i];size[col]++;
		for(int j=0;j<vec[i].size();j++){
			if(tar[vec[i][j]]!=col)
				outdu[col]++;
		}
	}
	for(int i=1;i<=Col;i++){
		if(!outdu[i])
			ans+=size[i],anst++;
	}
	if(anst>1) printf("0
");
	else printf("%d
",ans);
}