POJ 3660 Cow ContestCow(Floyd传递闭包)题解

题意:给出m个关系,问你能确定机头牛的排名

思路:要确定排名那必须要把他和其他n-1头牛比过才行,所以Floyd传递闭包,如果赢的+输的有n-1就能确定排名。

代码:

#include<cstdio>
#include<set>
#include<map>
#include<cmath>
#include<stack>
#include<vector>
#include<queue>
#include<cstring>
#include<string>
#include<sstream>
#include<iostream>
#include<algorithm>
#define ll long long
using namespace std;
const int maxn = 100+10;
const int INF = 0x3f3f3f3f;
int mp[maxn][maxn],can[maxn];
int main(){
    int n,m;
    memset(mp,0,sizeof(mp));
    memset(can,0,sizeof(can));
    scanf("%d%d",&n,&m);
    while(m--){
        int u,v;
        scanf("%d%d",&u,&v);
        mp[u][v] = 1;
    }
    for(int k = 1;k <= n;k++){
        for(int i = 1;i <= n;i++){
            for(int j = 1;j <= n;j++){
                if(mp[i][k] && mp[k][j]){
                    mp[i][j] = 1;
                }
            }
        }
    }
    int ans = 0;
    for(int i = 1;i <= n;i++){
        for(int j = 1;j <= n;j++){
            if(mp[i][j] && i != j){
                can[i]++;
                can[j]++;
            }
        }
    }
    for(int i = 1;i <= n;i++){
        if(can[i] == n - 1) ans++;
    }
    printf("%d\n",ans);
    return 0;
}

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转载自www.cnblogs.com/KirinSB/p/9486507.html