Description

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 cow A has a greater skill level than cow B (1 ≤ A ≤ N; 1 ≤ B ≤ N; A ≠ B), then cow A will always beat cow B.

Farmer John is trying to rank the cows by skill level. Given a list the results of M (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 5

4 3

4 2

3 2

1 2

2 5

Sample Output

2

Source

USACO 2008 January Silver

题目大意：给定N个人，然后知道这N个人的两两对决的情况，前一个为胜者，问M个对决情况中最多能确定几个人的名次.

思路：想到了Floyd算法，可是怎么来关联还是不知所以然，网上查了资料后，才明白用Floy求传递闭包，只要该人的关系和其他人的关系确定，那么他的名次就是确定的。

Code

#include <iostream>#include <cstring>#include <algorithm>using namespace std;

const int N=110;int mp[N][N];

int main(){    int n,m;    while (cin>>n>>m)    {        int i,j,k;        int u,v;        memset(mp,0,sizeof(mp));        for (i=0;i<m;i++)        {            cin>>u>>v;            mp[u][v]=1;        }       for(k=1;k<=n;k++)          for(i=1;i<=n;i++)              for(j=1;j<=n;j++)                if(mp[i][k]==1&&mp[k][j]==1)  //A和B有关，B和C有关，则Ac有关，不等号的传递性                    mp[i][j]=1;        //Floyd();        int ans=0;        for (i=1;i<=n;i++)        {            for (j=1;j<=n;j++)            {                if (j==i) continue;            if (mp[i][j]==0&&mp[j][i]==0) break;            }            if (j>n) ans++;        }        cout<<ans<<endl;    }}