POJ 3660—— Cow Contest——————【Floyd传递闭包】

Cow Contest

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 3660

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

题目大意：有n头牛，m个击败关系。问你最后有多少头牛的名次是可以确定的。

解题思路：Floyd传递闭包后，判断牛i前面有多少头牛，他后边有多少头牛。如果前后牛的头数等于n-1，那么说明他是可以确定名次的。

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
using namespace std;
int d[300][300];
int main(){
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF){
        int a,b;
        for(int i = 0; i < m;i++){
            scanf("%d%d",&a,&b);
            d[a][b] = 1;
        }
        for(int k = 1; k <= n; k++){
            for(int i = 1; i <= n; i++){
                for(int j = 1; j <= n; j++){
                    d[i][j] = (d[i][j]|| (d[i][k]&&d[k][j]));
                }
            }
        }
        int res = 0;
        for(int i = 1; i <= n; i++){
            int num = 0;
            for(int j = 1; j <=n; j++){
                if(j == i) continue;
                if(d[i][j] || d[j][i]) num++;
            }
            if(num == n-1) res++;
        }
        printf("%d\n",res);
    }
    return 0;
}