POJ3660 Cow Contest(floyd)

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

给出n头牛，m组数据。每一组数据前面的牛比后面的牛厉害。问你可以决定排名牛的数量。

floyd即可，如果可以确定排名则tmp等于n - 1。

AC代码：

#include "iostream"
#include "cstdio"
#include "cstring"
#include "algorithm"
using namespace std;
const int maxn = 105;
int n, m, map[maxn][maxn];
void floyd()
{
	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]) map[i][j] = 1;
}
int main(int argc, char const *argv[])
{
	while(scanf("%d%d", &n, &m) != EOF) {
		memset(map, 0, sizeof(map));
		for(int i = 0; i < m; ++i) {
			int a, b;
			scanf("%d%d", &a, &b);
			map[a][b] = 1;
		}
		floyd();
		int ans = 0;
		for(int i = 1; i <= n; ++i) {
			int tmp = 0;
			for(int j = 1; j <= n; ++j)
				tmp += (map[i][j] + map[j][i]);
			if(tmp == n - 1) ans++;			
		}
		printf("%d\n", ans);
	}
	return 0;
}