Cow_Contest

最新推荐文章于 2022-12-28 19:31:32 发布

Louzen

最新推荐文章于 2022-12-28 19:31:32 发布

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分类专栏：算法

本文链接：https://blog.youkuaiyun.com/Louzen/article/details/46540191

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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

#include
#include
#include
#include
#include
#include
using namespace std;

int n,m;
int x,y;
int ans;
bool h;
int map[110][110];

int main()
{
    freopen("C://Users//Lenovo//Desktop//in.txt","r",stdin);
    while(scanf("%d%d",&n,&m)!=EOF){
        ans=0;
        memset(map,0,sizeof(map));
        while(m--){
            scanf("%d%d",&x,&y);
            map[x][y]=1;
        }
        for(int i=1;i<=n;++i){
            for(int j=1;j<=n;++j){
                for(int k=1;k<=n;++k){
                    if(map[j][i]&&map[i][k])
                        map[j][k]=1;
                }
            }
        }
        for(int i=1;i<=n;++i){
            h=true;
            for(int j=1;j<=n;++j){
                if(i==j)
                    continue;
                if(map[i][j]==0&&map[j][i]==0){
                    ++ans;
                    break;
                }
            }
        }
        printf("%d\n",n-ans);
    }
    return 0;
}
注意Floyd中是if(map[j][i]&&map[i][k])就怎样怎样，而不是if(map[i][j]&&map[j][k])就怎样怎样，要把中间的那个点的遍历放在第一个for语句中，不知道为什么，放在下面的for语句里就会出错