nyoj 211&&poj 3660

最新推荐文章于 2017-04-29 16:27:05 发布

weixin_30672019

最新推荐文章于 2017-04-29 16:27:05 发布

阅读量67

点赞数

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/nyist-xsk/p/7264868.html

本文解析了CowContest编程竞赛题目，通过Floyd传递闭包算法确定参赛奶牛的技能排名，给出了完整的代码实现。

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

////能打败的个数加上被打败的个数恰好等于n-1,则能确定,
////否则无法确定,抽象为简单的floyd传递闭包算法,
////在加上每个顶点的出度与入度 (出度+入度=顶点数-1,则能够确定其编号)。
#include<queue>
#include<stack>
#include<vector>
#include<math.h>
#include<stdio.h>
#include<numeric>//STL数值算法头文件
#include<stdlib.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<functional>//模板类头文件
using namespace std;

const int INF=1e9+7;
const int maxn=102;

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

转载于:https://www.cnblogs.com/nyist-xsk/p/7264868.html