hdu 1217 汇率 Floyd

题意：给几个国家，然后给这些国家之间的汇率。判断能否通过这些汇率差进行套利交易。

Floyd的算法可以求出任意两点间的最短路径，最后比较本国与本国的汇率差，如果大于1，则可以。否则不可以。

有向图 一个点到另一点的花费为权值相乘 求乘积的最大值 从点i出发 再回到点i的花费如果大于1 就可以

Sample Input
3
USDollar
BritishPound
FrenchFranc
3
USDollar 0.5 BritishPound
BritishPound 10.0 FrenchFranc
FrenchFranc 0.21 USDollar

3
USDollar
BritishPound
FrenchFranc
6
USDollar 0.5 BritishPound
USDollar 4.9 FrenchFranc
BritishPound 10.0 FrenchFranc
BritishPound 1.99 USDollar
FrenchFranc 0.09 BritishPound
FrenchFranc 0.19 USDollar

0

Sample Output
Case 1: Yes
Case 2: No

 # include <iostream>
 # include <cstdio>
 # include <cstring>
 # include <string>
 # include <algorithm>
 # include <cmath>
 # include <map>
 # define LL long long
 using namespace std ;

 map<string,int>name;
 const int MAXN =  ;
 double dis[MAXN][MAXN];
 int n ;

 void floyed()//节点从1~n编号
 {
     int i,j,k;
     for(k=;k<=n;k++)
        for(i=;i<=n;i++)
          for(j=;j<=n;j++)
              if(dis[i][k]*dis[k][j] > dis[i][j])
                  dis[i][j]=dis[i][k]*dis[k][j];

 }

 int main()
 {
     //freopen("in.txt","r",stdin) ;
     int i,m,j;
     string str1,str2;
     double w;
     int iCase=;
     while(scanf("%d",&n),n)
     {
         iCase++;
         for(i=;i<=n;i++)
         {
             cin>>str1 ;
             name[str1]=i;
         }
         for(i=;i<=n;i++)
            for(j=;j<=n;j++)
            {
                if(i==j)dis[i][j]=;
                else dis[i][j]=;
            }
         scanf("%d",&m);
         while(m--)
         {
             cin>>str1>>w>>str2;
             dis[name[str1]][name[str2]]=w;
         }
         floyed();
         bool flag=false;
         for(i=;i<=n;i++)
           if(dis[i][i]>)
           {flag=true;break;}
         if(flag)  printf("Case %d: Yes\n",iCase);
         else printf("Case %d: No\n",iCase);
     }
     return ;
 }