POJ - 3660 - Cow Contest

- 日理万妓 2022-06-12 12:25 195阅读 0赞

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题目连接：[http://poj.org/problem?id=3660][http_poj.org_problem_id_3660]

## 题目描述 ##

### 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.

有n(1<=n<=100)个学生参加编程比赛。

给出m条实力信息。(1<=M<=4500)

其中每一条的格式为 A B （1<=A<=N,1<=B<=N,A!=B) 意思是A的实力比B强。

如果A比B强且B比C强，那么A一定比C强。

问最后有多少名学生可以确定他的排名。

保证输入信息不存在矛盾

### 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

第一行n和m。

以下m行 A B 表示A实力比B强。

### Output ###

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

输出答案

### Sample Input ###

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

### Sample Output ###

## 解题思路 ##

对每给的一个胜负关系连一条边，最后跑一次Floyd，然后判断一头牛所确定的关系是否是n-1次，若是，则这头牛的排名可以确定  
简单说就是这头牛可以和另一只不管怎样如果能连接到，就说明时是可以确定的

## AC代码 ##

#include<iostream>
    #include<cstring>
    using namespace std;
    int dis[110][110];
    int main () {
        int n, m;
        int a, b;
        while(~scanf("%d %d", &n, &m)) {
            memset(dis, 0, sizeof(dis));
            for(int i = 1; i <= m; i++) {
                cin >> a >> b;
                dis[a][b] = 1;
            }
            for(int k = 1; k <= n; k++) {
                for(int i = 1; i <= n; i++) {
                    for(int j = 1; j <= n; j++) {
                        dis[i][j] = dis[i][j] || (dis[i][k] && dis[k][j]);
                    } 
                }
            }
            int ans = 0;
            for(int i = 1; i <= n; i++) {
                int sum = 0;
                for(int j =1; j <= n; j++) {
                    if(dis[i][j] || dis[j][i])
                        sum++;
                }
                if(sum == n-1) 
                    ans++;
            }
            printf("%d\n", ans);
        }
        return 0;
    }

[http_poj.org_problem_id_3660]: http://poj.org/problem?id=3660