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  • 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
    ",ans);
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/KirinSB/p/9486507.html
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