POJ - 3660 Cow Contest

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最新推荐文章于 2022-03-14 21:26:15 发布

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分类专栏： ACM-图论-最短路

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本文介绍了一种解决牛牛编程比赛排名问题的算法。通过分析比赛结果，使用Floyd算法来确定每头牛的胜败关系，进而找出能够确切排名的牛的数量。

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Cow Contest

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 13409		Accepted: 7465

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

Sample Output

Source

USACO 2008 January Silver

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题意：有n头牛比赛，m个比赛结果，最后问你一共有多少头牛的排名被确定了；

其中如果a战胜b，b战胜c，则也可以说a战胜c，即可以传递胜负。求能确定排名的牛的数目。

思路：用Floyd判断连通，失败和胜利场数等于n-1即确定名次

Ac code:

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <stack>
#include <string>
#include <map>
using namespace std;
#define LL long long
#define INF 0x3f3f3f3f
#define fi first
#define se second
#define eps 1
const int mod=1000000000+7;
const int maxn=4500+5;

int dis[maxn][maxn];

void Floyd(int n){
    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 main() {
    int n,m;
    while(~scanf("%d%d",&n,&m)){
        memset(dis,0,sizeof dis);
        //for(int i=1;i<=n;i++) dis[i][i]=0;
        for(int i=1;i<=m;i++){
            int a,b;
            scanf("%d%d",&a,&b);
            dis[a][b]=1;
        }
        Floyd(n);
        int cnt,ans,res;
        cnt=ans=res=0;
        for(int i=1;i<=n;i++){
            ans=0;
            for(int j=1;j<=n;j++){
                if(dis[i][j]||dis[j][i]) ans++;
            }
            if(ans==n-1) res++;
        }
        printf("%d\n",res );

    }

}