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 = 35 ;
double dis[MAXN][MAXN];
int n ;

void floyed()//节点从1~n编号
{
    int i,j,k;
    for(k=1;k<=n;k++)
       for(i=1;i<=n;i++)
         for(j=1;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=0;
    while(scanf("%d",&n),n)
    {
        iCase++;
        for(i=1;i<=n;i++)
        {
            cin>>str1 ;
            name[str1]=i;
        }
        for(i=1;i<=n;i++)
           for(j=1;j<=n;j++)
           {
               if(i==j)dis[i][j]=1;
               else dis[i][j]=0;
           }
        scanf("%d",&m);
        while(m--)
        {
            cin>>str1>>w>>str2;
            dis[name[str1]][name[str2]]=w;
        }
        floyed();
        bool flag=false;
        for(i=1;i<=n;i++)
          if(dis[i][i]>1)
          {flag=true;break;}
        if(flag)  printf("Case %d: Yes
",iCase);
        else printf("Case %d: No
",iCase);
    }
    return 0;
}