No.9 - POJ2240 Floyd求负权值环路

Floyd求环：

// ShellDawn
// POJ2240
// No.9

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#include<iostream>
#include<map>
#define MM(x) memset(x,0,sizeof(x))
using namespace std;

#define maxn 35
int n;
map<string,int> mp;
double E[maxn][maxn];

int main(){
    int K = 1;
    while(~scanf("%d",&n)&&n!=0){
        mp.clear();
        for(int i=1;i<=n;i++){
            string s;
            cin>>s;
            mp[s] = i;
            E[i][i] = 1.0;
        }
        int T;
        scanf("%d",&T);
        for(int i=0;i<T;i++){
            string s1,s2;
            double v;
            cin>>s1>>v>>s2;
            int a = mp[s1];
            int b = mp[s2];
            E[a][b] = v;
        }
        for(int k=1;k<=n;k++){
            for(int i=1;i<=n;i++){
                for(int j=1;j<=n;j++){
                    E[i][j] = max(E[i][j],E[i][k]*E[k][j]);
                }
            }
        }
        bool flag = false;
        for(int i=1;i<=n;i++){
            if(E[i][i] > 1){
                flag = true;
                break;
            }
        }
        if(flag) printf("Case %d: Yes\n",K++);
        else printf("Case %d: No\n",K++);
    }
    return 0;
}