Arbitrage

描述

Arbitrage is the use of discrepancies in currency exchange rates to transform one unit of a currency into more than one unit of the same currency. For example, suppose that 1 US Dollar buys 0.5 British pound, 1 British pound buys 10.0 French francs, and 1 French franc buys 0.21 US dollar. Then, by converting currencies, a clever trader can start with 1 US dollar and buy 0.5 * 10.0 * 0.21 = 1.05 US dollars, making a profit of 5 percent.
Your job is to write a program that takes a list of currency exchange rates as input and then determines whether arbitrage is possible or not.

输入

The input file will contain one or more test cases. Om the first line of each test case there is an integer n (1<=n<=30), representing the number of different currencies. The next n lines each contain the name of one currency. Within a name no spaces will appear. The next line contains one integer m, representing the length of the table to follow. The last m lines each contain the name ci of a source currency, a real number rij which represents the exchange rate from ci to cj and a name cj of the destination currency. Exchanges which do not appear in the table are impossible.
Test cases are separated from each other by a blank line. Input is terminated by a value of zero (0) for n.

输出

对于每一个测试用例,打印一行告诉是否套利是可能的格式”案例案例:是的“分别”案例情况:没有。

样例输入

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

样例输出

Case 1: Yes
Case 2: No

代码

#include
#include
#include
using namespace std;
map<string,int>name;
#define MAXN 30
double g[MAXN][MAXN];
//Floyed算法,求任意两个顶点间的最短路径
//dis[][]记录任意两点间的最短路径，初始的dis[][]记录直接路径 
#define typec double
void floyed(typec dis[][MAXN],int n)//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][j]int main()
{
    int n,i,m,j;
    string str1,str2;
    double r;
    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)g[i][j]=1;
               else g[i][j]=0;
           }    
        scanf("%d",&m);   
        while(m--)
        {
            cin>>str1>>r>>str2;
            g[name[str1]][name[str2]]=r;
        }     
        floyed(g,n);
        bool flag=false;
        for(i=1;i<=n;i++)
          if(g[i][i]>1)
          {flag=true;break;}
        if(flag)  printf("Case %d: Yes\n",iCase);
        else printf("Case %d: No\n",iCase);
    }    
    return 0;
}   

测试截图

AC截图