Arbitrage HDU

Arbitrage

                                             Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
                                              Total Submission(s): 3180    Accepted Submission(s): 1445

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

 

Source

University of Ulm Local Contest 1996

 

 

         
只能感叹外国就连ACM比赛都是这么符合实际且实用啊。。。。。

题意：

        意思就是叫你算出银行的汇率，判断是否可以赚钱。。哈哈，当然这是非专业的。

        其实题目是告诉我们几个国家之间的汇率，然后叫我们判断他们之间是否有环。且最终通过这种转换能否赚钱。

 其实就是一个有关DP问题于是运用暴力的floyd就可以ko了。

#include <iostream>
#include <algorithm>
#include <string>
#include <set>
#include <map>
#include <cstdio>
#include <cstring>
using namespace std;

double dp[31][31];
bool Floyd(int n)
{
    for(int k = 1;k <= n;k++)
    for(int i = 1;i <= n;i++)
      for(int j = 1;j <= n;j++)
        dp[i][j] = max(dp[i][j],dp[i][k]*dp[k][j]);
    for(int i = 1;i <= n;i++)
     if(dp[i][i] > 1)
       return true;
    return false;
}

int main()
{
    double rate;
    int n,m,i,kase = 1;
    while(cin>>n,n)
    {
        string s,a,b;
        map<string,int> mp;
        memset(dp,0,sizeof(dp));
        for(i = 1;i <= n;i++)
        {
            cin>>s;
            mp[s] = i;
        }
        cin>>m;
        for(i = 1;i <= m;i++)
        {
            cin>>a>>rate>>b;
            dp[mp[a]][mp[b]] = rate;
        }
        if(Floyd(n))
          printf("Case %d: Yes\n",kase++);
        else
          printf("Case %d: No\n",kase++);
    }
    return 0;
}