POJ 3660 Cow Contest

最新推荐文章于 2020-12-19 00:08:56 发布

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最新推荐文章于 2020-12-19 00:08:56 发布

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分类专栏：图论算法

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本文介绍了一个编程竞赛中通过胜负记录确定参赛牛的技能排名的方法。使用Floyd算法更新胜者和败者的潜在对手关系，最终统计能够确切确定排名的参赛者数量。

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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 cowA has a greater skill level than cowB (1 ≤ A ≤ N; 1 ≤B ≤ N; A ≠B), then cow A will always beat cowB.

Farmer John is trying to rank the cows by skill level. Given a list the results ofM (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

Sample Output

一开始题意理解的有问题，上网找了个题解看了一下才弄明白题意，如果一个牛的排名可以确定，那么排在他前面的和排在他后面的牛的总数和为n-1，水题。。。附上代码：

#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
const int maxn = 100 + 5;
const int INF = 1 << 30;
bool a[maxn][maxn];
int n,m,sum;
void floyd()
{
    for(int k = 1;k <= n; ++k)
    {
        for(int i = 1;i <= n; ++i)
        {
            for(int j = 1;j <= n; ++j)
            {
                if(i == j || i == k || j == k) {continue;}
                if(a[i][k] && a[k][j])
                    a[i][j] = a[i][k] & a[k][j];
            }
        }
    }
}
int main()
{
    memset(a,false,sizeof(a));
    int x,y;
    cin>>n>>m;
    for(int i = 0;i < m; ++i)
    {
        cin >> x >> y;
        a[x][y] = true;
    }
    floyd();
    int sum = 0;
    for(int i = 1;i <= n; ++i)
    {
        x = y = 0;
        for(int j = 1;j <= n; ++j)
        {
            if(i == j) continue;
            if(a[i][j]) ++x;
            if(a[j][i]) ++y;
        }
        if(x + y == n-1) ++sum;
    }
    cout << sum << endl;
}