floyd(传递闭包)Cow Contest

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

题目给出了m对的相对关系，求有多少个排名是确定的。

使用floyed求一下传递闭包。如果这个点和其余的关系都是确定的，那么这个点的排名就是确定的。

 

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
#define inf 0x3f3f3f3f
int map1[5000][5000];
int main()
{
    int k,i,j,n,m;
    while(~scanf("%d%d",&n,&m))
    {
        int a,b;
        for(i=1; i<=m; i++)
        {
            scanf("%d%d",&a,&b);
            map1[a][b]=1;
        }
        for(k=1; k<=n; k++)
            for(i=1; i<=n; i++)
                for(j=1; j<=n; j++)
                    if(map1[i][k]==1&&map1[k][j]==1)
                        map1[i][j]=1;
        int sum=0;
        for(i=1; i<=n; i++)
        {
            for(j=1; j<=n; j++)
            {
                if(i==j)
                    continue;
                if(map1[i][j]==0&&map1[j][i]==0)
                    break;
            }
            if(j>n)
                sum++;
        }
        printf("%d\n",sum);
    }
    return 0;
}