poj 3660 floyd 传递闭包

poj 3660 Cow Contest

题目：

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

题意：

就是给你一堆大牛，给出一些大牛之间的比较关心(优劣)，形成一个有向无环图，问你有多少个大牛的排名是能确定的。

思路：

想一下，如果知道一个大牛前面有多少人，后面有多少人，那么他的排名一定是确定的，抽象到图上，就是对于一个排名确定的大牛，其他的大牛一定与他相连（或正向，或反向）。

代码：

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#define inf 0x3f3f3f3f
int n,m;
int maps[120][120];
void floyd()
{
for(int k=1;k<=n;k++)
for(int i=1;i<=n;i++)
{
if(maps[i][k]==inf)
continue;
for(int j=1;j<=n;j++)
{
if(maps[i][j]>maps[i][k]+maps[k][j])
maps[i][j]=maps[i][k]+maps[k][j];
}
}
int sum=0;
for(int i=1;i<=n;i++)
{
bool flag=true;
for(int j=1;j<=n;j++)
{
if(i==j||maps[i][j]!=inf||maps[j][i]!=inf)
continue;
else
{

97eb
flag=false;
break;
}
}
if(flag)sum++;
}
printf("%d\n",sum);
}
int main()
{
while(scanf("%d%d",&n,&m)!=EOF)
{
int x,y;
memset(maps,inf,sizeof(maps));
for(int i=1;i<=m;i++)
{
scanf("%d%d",&x,&y);
maps[x][y]=1;
}
floyd();
}
}