nyoj 211 Cow Contest

他说少年如歌

于 2017-05-05 11:24:38 发布

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本文介绍了一道关于编程竞赛排名确定的算法题。题目中N头技能各不相同的牛进行多轮比赛，根据比赛结果确定能够精确排名的牛的数量。通过Floyd算法实现胜者传递性质，最终统计出确定排名的牛的数量。

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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

样例输出

来源

USACO 2008 January Silver

题意：给你n头牛，每头牛都有自己的kills值，和其他牛的都不同，然后kills大的可以赢小的。现在给出n头牛的比赛结果，问你有几头牛可以确定排名。

如果A>B，B>C,那么A>C,所以这道题我们要用到floyd的闭包传递。

AC代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <cstring>
#define INF 0x3f3f3f3f
using namespace std;
int map[101][101];

int main()
{
    int n, m;
    while(~scanf("%d%d",&n,&m))
    {
        if(n==0&&m==0)
            break;
        memset(map, 0, sizeof(map));
        for(int i = 1; i <= m; i++)
        {
            int x, y;
            scanf("%d%d",&x,&y);
            map[x][y] = 1;
        }
        for(int k = 1; k <= n; k++)
        {
            for(int i = 1; i <= n; i++)
            {
                for(int j = 1; j <= n; j++)
                {
                    if(map[i][k]&&map[k][j])
                    {
                        map[i][j] = 1;
                    }
                }
            }
        }
        int c1 = 0, c2 = 0;
        int ans = 0;
        for(int i = 1; i <= n; i++)
        {
            c1 = 0, c2 = 0;
            for(int j = 1; j <= n; j++)
            {
                if(map[i][j])
                    c1++;
                if(map[j][i])
                    c2++;
            }
            if(c1+c2==n-1)
                ans++;
        }
        printf("%d\n",ans);
    }
    return 0;
}