Arbitrage （HDU 1217）

                       Arbitrage

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4160    Accepted Submission(s): 1890

Problem Description

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.

Input

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.

Output

For each test case, print one line telling whether arbitrage is possible or not in the format "Case case: Yes" respectively "Case case: No".

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
   
   
   
   

这题只要理解了最短路的解法思想，就不难AC了。

#include<iostream>
#include<string>
#include<map>
using namespace std;
const int INF = 0.00001; //由于求最大值，所以应该用INF来表示无限小的数
int main()
{
    int n,m,i,j,k,t=0;
    double d[50][50];
    string x[50],x1,x2;

    while(cin >> n, n)
    {
        for(i=1; i<=n; i++)
            for(j=1; j<=n; j++)
                d[i][j] = INF;
        map<string,int> p;
        for(i=1; i<=n; i++)
        {
            cin >> x[i];
            p[x[i]] = i;
        }
        cin >> m;
        double c;
        for(i=1; i<=m; i++)
        {
            cin >> x1 >> c >> x2;
            d[p[x1]][p[x2]] = c;
        }
        for(k=1; k<=n; k++)
          for(i=1; i<=n; i++)
            for(j=1; j<=n; j++)
               if(d[i][k] * d[k][j] > d[i][j]) //求最大，所以是">".
                  d[i][j] = d[i][k] * d[k][j];
        cout << "Case " << ++t << ": ";
        //cout << d[1][1] << endl;
        if(d[1][1] > 1) cout<<"Yes"<<endl; //由于是求自己到自己的转换，所以是d[1][1];
        else  cout<<"No"<<endl;           //如果是求起点到终点n的转换，就应该是d[1][n];
    }
    return 0;
}