FJUT 1521Cow Contest

最新推荐文章于 2019-08-12 00:22:15 发布

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#暴力

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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

SampleInput
5 5
4 3
4 2
3 2
1 2
2 5
SampleOutput
2

给你n头牛 m种关系每行 x y 表示 x排名大于y 问你有多少头牛的排名是确定的。。看网上说什么闭包拓扑排序啊啥的其实数据范围那么小暴力大法就好了

#include<iostream>
using namespace std;
int main()
{

    int v[105][105]= {0};
    int n,m;
    cin>>n>>m;
    while(m--)
    {
        int x,y;
        cin>>x>>y;
        v[y][x]=1;

    }
    for(int i=1; i<=n; i++)
        for(int j=1; j<=n; j++)
            for(int k=1; k<=n; k++)
                if(v[j][i]&&v[i][k])
                    v[j][k]=1;
    int sum=0;
    for(int i=1; i<=n; i++)
    {
        int s=0;
        for(int j=1; j<=n; j++)
            s+=v[i][j]+v[j][i];

        if(s==n-1)
            sum++;
    }
    cout<<sum<<endl;
}

解释下代码吧 v【i】【j】 =1 说明i的排名小于j 然后比如有v【i】【j】==1&&v【j】【k】==1
那么是不是一定有v【i】【k】==1
思考一下我们要知道这头牛的排名是不是一定要知道其他牛跟他的排名关系是大于还是小于所以我们如果想知道牛x的排名只需要遍历加v【x】【j】和v【j】【x】当两个都为0时候说明x与j的关系不确定谁排名靠前有确切的关系就说明j排名大于他或者小于他也就是v值等于1 当sum满足n-1时候说明牛x的排名是确切的