POJ 2240 Arbitrage

/* bellman，加上map，判断是否有负权回路即可
* */

#include < iostream >
#include < string >
#include < map >
using   namespace  std;
#define  MAX 1000
#define  INF 999999
#define  EPS 1e-8
struct  Node{
     int  u,v;
     double  val;
}node[MAX];
int  nv,ne,start;
double  dist[MAX];
int  bellman()
{
     int  i,j,flag;
     double  tmp;
     for (i = 1 ;i <= nv; ++ i)
        dist[i]  =   - 1 ;
    dist[start]  =   1 ;
     for (i = 1 ;i < nv; ++ i)
    {
        flag  =   0 ;
         for (j = 1 ;j <= ne; ++ j)
        {
             if (dist[node[j].u] == INF)
                 continue ;
            tmp  =  dist[node[j].u] * node[j].val;
             if (tmp > dist[node[j].v] + EPS)
            {
                dist[node[j].v]  =  tmp;
                flag  =   1 ;
            }
        }
         if (flag  ==   0 )  return   0  ;
        
    }
     for (i = 1 ;i <= ne; ++ i)
         if (dist[node[i].u] != INF && dist[node[i].u] * node[i].val > dist[node[i].v] + EPS)
             return   1 ;
     return   0 ;
}
int  main()
{
     int  i,j,cnt = 0 ;
     double  val;
     char  a[ 100 ],b[ 100 ];
    map < string , int >  name;
     while ( 1 )
    {
        scanf( " %d " , & nv);
         if ( ! nv)  break ;
        name.clear();
         int  flag  = 0  ;
         ++ cnt;
         for (i = 1 ;i <= nv; ++ i)
        {
            scanf( " %s " ,a);
            name[ string (a)]  =  i;
        }
        scanf( " %d " , & ne);
         for (i = 1 ;i <= ne; ++ i)
        {
            scanf( " %s%lf%s " , & a, & val, & b);
            node[i].u  =  name[ string (a)];
            node[i].v  =  name[ string (b)];
         //     if(node[i].v==node[i].u&&val>1)
         //         flag = 1;
            node[i].val  =  val;
        }
        start  =   1 ;
        printf( " Case %d:  " ,cnt);
         if (flag  ==   1 )
            printf( " Yes\n " );
         else   if (bellman())
            printf( " Yes\n " );
         else
            printf( " No\n " );
    }
     return   0 ;
}