poj 2240 Arbitrage

这个题求是否经过兑换后钱能不能增加。

这题可以用bellman比较容易解决；

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<queue>
#include<set>
#include<cstring>
#include<string>
#include<vector>
using namespace std;
class Edge
{
    public:
        int S,E;
        double rate;    
}edge[1024];
bool bell_man( int N ,int M )
{
    double dis[50];
    for( int i = 0 ; i <= N ;i ++ )
         dis[i] = 1.0;
    int flag = 0;
    for( int i = 0; i < N +4 ; i++ )
    {
        flag = 0;
        for( int j =  0; j < M ;j ++ )
        {
           if( dis[edge[j].E] < dis[edge[j].S]*edge[j].rate )
           {
               dis[edge[j].E] = dis[edge[j].S]*edge[j].rate;
               flag = 1;
           }    
        }
        if( !flag ) return false; 
    }
    return true;
}
int main(  )
{
    string str[50],start,end;
    int N,M,S,E,Case=1;
    double rate;
    while( scanf( "%d",&N ),N )
    {
        for( int i = 1 ; i <= N; i ++ )
        {
            cin>>str[i];    
        }
        scanf( "%d",&M );
        for( int i =0 ; i < M ; i ++ )
        {
            cin>>start>>rate>>end;
            for(int j = 1 ; j <= N ; j++)
            {
                if( start == str[j] ) S = j;
                if( end == str[j] ) E = j;    
            }    
            edge[i].S= S ;edge[i].E = E;
            edge[i].rate = rate;
        }
        printf( "Case %d: ",Case++ );
        if( bell_man(N , M) ) printf( "Yes\n" );
        else printf( "No\n" );     
    }
    //system( "pause" );
    return 0;
}