# POJ3660 Cow Contest(Floyd,传递闭包)

Cow Contest
 Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 10767 Accepted: 5996

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 ≤ NA ≠ 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

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#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <string>
#include <iostream>
#include <stack>
#include <queue>
#include <vector>
#include <algorithm>
#define mem(a,b) memset(a,b,sizeof(a))
#define N 100+20
#define M 100000+20
#define inf 0x3f3f3f3f
using namespace std;
int n,m;
int map[N][N];
int main()
{
int a,b;
while(~scanf("%d%d",&n,&m))
{
mem(map,0);
for(int i=1; i<=m; i++)
{
scanf("%d%d",&a,&b);
map[a][b]=1;
}
int sum=0;
for(int k=1; k<=n; k++)
for(int i=1; i<=n; i++)
for(int j=1; j<=n; j++)
if(map[i][k]&&map[k][j])//间接变成直接
map[i][j]=1;

for(int i=1; i<=n; i++)
{
int k=0;
for(int j=1; j<=n; j++)
k+=map[i][j]+map[j][i];
if(n-1==k)//如果一头牛和其他所有牛的关系确定了的话，它的排名也就确定了。
sum++;
}
printf("%d\n",sum);
}
return 0;
}  

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