poj-3660 Cow Contest（传递闭包floyed算法）

http://poj.org/problem?id=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

说明

Cow 2 loses to cows 1, 3, and 4. Thus, cow 2 is no better than any of the cows 1, 3, and 4. Cow 5 loses to cow 2, so cow 2 is better than cow 5. Thus, cow 2 must be fourth, and cow 5 must be fifth. The ranks of the other cows cannot be determined.

题意：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图，然后进行适当的松弛操作（使用floyed求一下传递闭包），得到任意两点的关系，再对每一头牛进行检查即可。

 

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <string>
#include <math.h>
#include <algorithm>
#include <queue>
const int INF=0x3f3f3f3f;
using namespace std;
int D[101][101];

int main()
{
    int n,m;
    scanf("%d %d",&n,&m);
    //输入信息 
    for(int i=1;i<=m;i++)
    {
        int a,b;
        scanf("%d %d",&a,&b);
        D[a][b]=1;//a等级大于b 
        D[b][a]=-1;//b等级小于a 
    }
    
    //进行适当的松弛操作（使用floyed求一下传递闭包）
    for(int k=1;k<=n;k++)
    {
        for(int i=1;i<=n;i++)
        {
            for(int g=1;g<=n;g++)
            {
                if(D[i][k]==1&&D[k][g]==1)
                    D[i][g]=1;
                else if(D[i][k]==-1&&D[k][g]==-1)
                    D[i][g]=-1;
                    
            }
        }
    }
    
    //统计答案 
    int ans=0;
    for(int i=1;i<=n;i++)
    {
        int g;
        for(g=1;g<=n;g++)
        {
            if(i!=g&&!D[i][g])//除i=g时，D[i][g]为0说明i对g无确定的关系 
                break;
        }
        if(g==n+1)//说明该牛对其余的牛都有确定的关系 
            ans++;
    }
    printf("%d
",ans);
    return 0;
}