最短路(Floyd_Warshall) POJ 2240 Arbitrage

题目传送门

 /*
     最短路：Floyd模板题
         只要把+改为*就ok了，热闹后判断d[i][i]是否大于1
     文件输入的ONLINE_JUDGE少写了个_，WA了N遍:)
 */
 #include <cstdio>
 #include <iostream>
 #include <cstring>
 #include <algorithm>
 #include <string>
 #include <map>
 #include <cmath>
 #include <vector>
 #include <set>
 #include <queue>
 using namespace std;

 const int MAXN = 1e6 + ;
 const int INF = 0x3f3f3f3f;
 double d[][];

 void Floyd_Warshall(int n)
 {
     for (int k=; k<=n; ++k)
     {
         for (int i=; i<=n; ++i)
         {
             for (int j=; j<=n; ++j)
             {
                 if (d[i][j] < d[i][k] * d[k][j])
                 {
                     d[i][j] = d[i][k] * d[k][j];
                 }
             }
         }
     }

     for (int i=; i<=n; ++i)
     {
         if (d[i][i] > )
         {
             puts ("Yes");    return ;
         }
     }

     puts ("No");    return ;
 }

 int main(void)        //POJ 2240 Arbitrage
 {
     #ifndef ONLINE_JUDGE
     freopen ("F.in", "r", stdin);
     #endif

     int n, num;    int cas = ;
     while (cin >> n && n)
     {
         for (int i=; i<=n; ++i)
         {
             for (int j=; j<=n; ++j)
             {
                 if (i == j)        d[i][j] = ;
                 else    d[i][j] = ;
             }
         }

         map<string, int> m;
         string s, s1, s2;
         for (int i=; i<=n; ++i)
         {
             cin >> s;
             m[s] = i;
         }

         cin >> num;
         for (int i=; i<=num; ++i)
         {
             double w;
             cin >> s1 >> w >> s2;
             d[m[s1]][m[s2]] = w;
         }

         cout << "Case " <<  ++cas << ": ";
         Floyd_Warshall (n);
     }

     return ;
 }