最短路径(8)--poj3660(用floyd求传递闭包)

Cow Contest

                                         Time Limit:1000MS    Memory Limit:65536KB    64bit IO Format:%lld & %llu

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

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

           

           这道题题目大意：有N头牛，评以N个等级，各不相同，先给出部分牛的等级的高低关系，问最多能确定多少头牛的等级。

解题思路：一头牛的等级，当且仅当它与其它N-1头牛的关系确定时确定，于是我们可以将牛的等级关系看做一张有向图，然后进行适当的松弛操作，得到任意两点的关系，再对每一头牛进行检查即可。即利用floyd求传递闭包。

代码：

#include 
#include 
#define MAX 105
int map[MAX][MAX];
int n,m;
void Floyd()
{
	int i,j,k;
	for(k=1;k<=n;k++)
	{
		for(i=1;i<=n;i++)
		{
			for(j=1;j<=n;j++)
			{
				if(!map[i][j])map[i][j]=(map[i][k]&&map[k][j]);
			}
		}
	}
}
int main()
{
	int i,j;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		memset(map,0,sizeof(map));
		for(i=0;i