[题解] poj 3660 Cow Contest (floyd)

- 传送门 -

　http://poj.org/problem?id=3660

#Cow Contest

| Time Limit: 1000MS |   | Memory Limit: 65536K |
| Total Submissions: 12726 |   | Accepted: 7084 |

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

[[Submit](http://poj.org/submit?problem_id=3660)]   [[Status](http://poj.org/problemstatus?problem_id=3660)]   [[Discuss](http://poj.org/bbs?problem_id=3660)]

- 题意 -

　给你 n 个点, m 个关于点权大小的判断, 例如 x y 表示 x 的权值大于 y.
　试判断能够在n个点中确定点权大小排名的点的个数.
　

- 思路 -

　对于 (x) 的权值大于 (y) , 不妨连一条有向边 (x o y), 那么点的入边(以及入边指向的点的入边...)表示比它大的点, 出边(以及出边指向的点的出边...)表示比它小的点, 如果我们能找到比一个点大和比它小的点加起来正好是 n - 1 个, 那么我们就能确定这个点的排名.
　发现跑一次 floyd 把间接联通的点直接连起来就可以了.
　注意单向边.
　
　细节见代码.
　

- 代码 -

#include<cstdio>
using namespace std;

int G[105][105];
int n, m;

void floyd() {
	for (int k = 1; k <= n; ++k)
		for (int i = 1; i <= n; ++i)
			for (int j = 1; j <= n; ++j) {
					if (i != j && j != k && k != i)
						G[i][j] = (G[i][j] || (G[i][k] && G[k][j]));
			}
}

int main() {
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= m; ++i) {
		int x, y;
		scanf("%d%d", &x, &y);
		G[x][y] = 1;
	}
	floyd();
	int ans = 0;
	for (int i = 1; i <= n; ++i) {
		int tmp = 1;
		for (int j = 1; j <= n; ++j)
			if (!G[i][j] && !G[j][i] && i != j) {
				tmp = 0;
				break;
			}
		ans += tmp;
	}
	printf("%d
", ans);
	return 0;
}