POJ3660CowContest
1048+

### 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 ≤ AN; 1 ≤ BN; AB), 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


# 解题思路

a能打败b，b能打败c，那么a就能打败c，这种打败关系是可以传递的，这就是传递闭包。用Floyd可以求传递闭包，在原来求多源最短路径的基础上，不求用最短路，而是能到就是1，不能到就是0。mp[i][j]表示i能打败j，那么i能打败多少头牛，就是mp[i]中1的个数，能被多少头牛打败，就是mp[][i]中1的个数，如果这两个加起来刚好是n-1，那么这头牛的排名就确定了。