nyoj 211——Cow Contest——————【floyd传递闭包】

Cow Contest

时间限制：1000 ms  |  内存限制：65535 KB

难度：4

 

描述

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.

 

输入

* 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

There are multi test cases.The input is terminated by two zeros.The number of test cases is no more than 20.

输出

For every case:
* Line 1: A single integer representing the number of cows whose ranks can be determined

样例输入

5 5
4 3
4 2
3 2
1 2
2 5
0 0

样例输出

2

题目大意：给你n，m表示n位大牛，有m对能力比较关系，表示a能打败b。问你最后几个人的排名可以确定。

解题思路：首先用floyd传递闭包，然后枚举统计排名可以确定的人数。某大牛的排名确定，则应该有他与其他n-1个人关系确定，败或赢。

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn=120;
int d[maxn][maxn];
void floy(int n){
    int i,j,k;
    for(k=1;k<=n;k++){
        for(i=1;i<=n;i++){
            for(j=1;j<=n;j++){
                d[i][j]=d[i][j]||(d[i][k]&&d[k][j]);
            }
        }
    }
}
int work(int n){
    int ret=0,sum,k,i,j;
    for(k=1;k<=n;k++){
        sum=0;
        for(i=1;i<=n;i++){
            if(i==k) continue;
            if(d[k][i]){
                sum++;
            }
            if(d[i][k]){
                sum++;
            }
        }
        if(sum==n-1)
            ret++;
    }
    return ret;
}
int main(){
    int n,m,i,j,k,a,b;
    while(scanf("%d%d",&n,&m)!=EOF&&(n+m)){
        memset(d,0,sizeof(d));
        for(i=0;i<m;i++){
            scanf("%d%d",&a,&b);
            d[a][b]=1;
        }
        floy(n);
        printf("%d\n",work(n)) ;
    }
    return 0;
}