题干：

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. 

题目大意：

有N(N<=10000)头牛，每头牛都想成为most poluler的牛，给出M(M<=50000)个关系，如(1,2)代表1欢迎2，关系可以传递，但是不可以相互，即1欢迎2不代表2欢迎1，但是如果2也欢迎3那么1也欢迎3.
给出N,M和M个欢迎关系，求被所有牛都欢迎的牛的数量。

解题报告：

  Tarjan缩点然后看集合数量是否是1，如果不是1那就输出0，如果是1，那就输出这个集合的数量。

AC代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
const int MAX = 2e5 + 5;
int n,m;
int head[MAX];
int DFN[MAX],LOW[MAX],col[MAX],cnt[MAX],stk[MAX],out[MAX];
bool vis[MAX];
struct Edge {
  int fr,to,ne;
} e[MAX],ee[MAX];
int tot,tot2,timing,scc,index;
void add(int u,int v) {
  e[++tot].fr = u;
  e[tot].to = v;
  e[tot].ne = head[u];
  head[u] = tot;
}
void Tarjan(int x) {
  LOW[x] = DFN[x] = ++timing;
  vis[x] = 1;
  stk[++index] = x;
  for(int i = head[x]; i!=-1; i=e[i].ne) {
    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],DFN[v]);
    }
  }
  if(DFN[x] == LOW[x]) {
    scc++;
    while(1) {
      int tmp = stk[index];index--;
      vis[tmp]=0;
      col[tmp] = scc;
      cnt[scc]++;
      if(x == tmp) break;
    }
  }
}
int main()
{
  cin>>n>>m;
  memset(head,-1,sizeof head);
  for(int a,b,i = 1 ; i<=m; i++) {
    scanf("%d%d",&a,&b);
    add(a,b);
  }
  for(int i = 1; i<=n; i++) {
    if(!DFN[i]) Tarjan(i);
  }
  for(int i = 1; i<=m; i++) {
    if(col[e[i].fr] != col[e[i].to]) {
      out[col[e[i].fr]]++;
    }
  }
//  for(int u = 1; u<=n; u++) {
//    for(int i = head[u]; i!=-1; i=e[i].ne) {
//      int v = e[i].to;
//      if(col[u] != col[v]) out[]
//    }
//  }
  int emmm=0,ans=0;
  for(int i = 1; i<=scc; i++) {
    if(out[i] == 0) {
      emmm++;ans = cnt[i];
    } 
  }
  if(emmm != 1) printf("0\n");
  else printf("%d\n",ans);
  return 0 ;
}