[floyd]poj3660 Cow Contest

最新推荐文章于 2021-09-08 19:31:03 发布

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最新推荐文章于 2021-09-08 19:31:03 发布

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分类专栏： 5、图论 ————ACM训练————

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————ACM训练———— 同时被 2 个专栏收录

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5、图论

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本文介绍了一种通过分析牛之间比赛结果来精确确定其技能排名的方法。利用传递关系更新每轮比赛的结果，确保最终能够准确识别出那些技能等级可以确切决定的参赛牛。

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Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

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

给出各个牛之间的强弱关系，判断多少个牛的排名能被确定下来。
用prim做，对于i,k和j，如果i>k且k>j则能说明i>j，修改所有的传递关系。
在数组内做完预处理后，对每个牛判断与各个的排名关系。如果这头牛和其他牛之间都有对应的胜负关系，那么就说明它的排名可以确定，加入计数器，统计完所有的牛以后再输出。

#include <queue>
#include <iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int MAXN=102;
int d[MAXN][MAXN];
int N,M;

int main()
{
    int a,b;
    while(cin>>N>>M)
    {
        memset(d,0,sizeof(d));
        for(int i = 0; i < M; i++)
        {
            cin>>a>>b;
            d[a][b] = 1;
        }
        for(int k = 1; k<=N;k++)
            for(int i = 1; i <= N; i++)
                for(int j = 1; j<=N; j++)
                    if(d[i][k] && d[k][j])
                        d[i][j]=1;
        int ans = 0;
        for(int i = 1; i<=N;i++)
        {
            int res = N-1;
            for(int j = 1;j<=N;j++)
            if(d[i][j] || d[j][i])
                res--;
            if (!res)
                ans++;
        }
        cout<<ans<<endl;
    }
}