Cow Contest

N (1 ≤ N ≤ 100) cows, conveniently numbered 1..N, are participating in a programming contest. As we all know, some cows code better than others. Each cow has a certain constant skill rating that is unique among the competitors.

The contest is conducted in several head-to-head rounds, each between two cows. If cowA has a greater skill level than cow B (1 ≤ A ≤ N; 1 ≤B ≤ N; A ≠ B), then cow A will always beat cowB.

Farmer John is trying to rank the cows by skill level. Given a list the results ofM (1 ≤ M ≤ 4,500) two-cow rounds, determine the number of cows whose ranks can be precisely determined from the results. It is guaranteed that the results of the rounds will not be contradictory.

Input

* Line 1: Two space-separated integers: N and M
* Lines 2..M+1: Each line contains two space-separated integers that describe the competitors and results (the first integer,A, is the winner) of a single round of competition: A and B

Output

* Line 1: A single integer representing the number of cows whose ranks can be determined
　

Sample Input

5 54 34 23 21 22 5

Sample Output

2

我们要想知道一个点的rank排名的话，我们必须知道他和剩余n-1的关系都确定之后，才能确定。我们用floyd来确定一个人和所有人的关系。

#include <cstdio>#include<cstring>#include<queue>#include<algorithm>using namespace std;typedef long long ll;int N,M;int mp[110][110];const int INF=0x3f3f3f3f;void floyd(){    for(int k=1;k<=N;k++){        for(int i=1;i<=N;i++){            for(int j=1;j<=N;j++){                if(mp[i][k]==mp[k][j]&&mp[i][k]!=INF){//如果说i->k的关系确定，k->j的关系确定的话，i->j的关系就确定了                    mp[i][j]=mp[i][k];                }            }        }    }}int main(){   scanf("%d %d",&N,&M);   memset(mp,INF,sizeof(mp));   for(int i=1;i<=M;i++){    int u,v;     scanf("%d %d",&u,&v);     mp[u][v]=1;     mp[v][u]=-1;   }   floyd();   int ans=0;   for(int i=1;i<=N;i++){     int sum=0;     for(int j=1;j<=N;j++){        if(mp[i][j]!=INF)            sum++;     }     if(sum==N-1)        ans++;   }   printf("%d\n",ans);   return 0;}