POJ 2240Arbitrage（最短路floyd）

Arbitrage

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 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

题意：通过各个国家的汇率比不同来赚钱，从一种货币出发最后回归到这种货币，例一：1美元=0.5英镑，1英镑=10法币，1法币=0.21美元，有1美元可以兑换0.5英镑，0.5英镑兑换5法币，5法币兑换1.05美元，从而赚了0.05美元。首先输入一些货币，然后是各种货币的汇率比。如果可以从中赚钱就输出Yes，否则输出No。

题解：1、通过Map容器来存各种货币。

2、用floyd算法。

3、找出本身的钱有没有变多。

#include <iostream>
#include <map>
#include <string.h>
using namespace std;
const int Ink = 0X3f3f3f3f;
int main()
{
  int n, m;
  double w;
  int cesa = 0;
  int i, j, k;
  char s[50], str1[50], str2[50];
  double Map[50][50];
  while(cin>>n)
    {
      if(n == 0)
        return 0;
      cesa ++;
      map<string, int>str;
      memset(Map, Ink, sizeof(Map));
      for(i = 1; i <= n; i++)
        {
          cin>>s;
          str[s] = i;
          Map[i][i] = 1;
        }
      cin>>m;
      for(i = 1; i <= m; i++)
        {
          cin>>str1>>w>>str2;
          Map[str[str1]][str[str2]] = w;
        }
      for(i = 1; i <= n; i++)
        {
          for(j = 1; j <= n; j++)
            {
              for(k = 1; k <= n; k++)
                {
                  if(Map[j][k] < Map[j][i] * Map[i][k])
                    {
                      Map[j][k] = Map[j][i] * Map[i][k];
                    }
                }
            }
        }
      int flag = 1;
      for(i = 1; i <= n; i++)
        {
          if(Map[i][i] > 1)
            {
              flag = 0;
              break;
            }
        }
      if(flag == 0)
        cout<<"Case "<<cesa<<": Yes"<<endl;
      else
        cout<<"Case "<<cesa<<": No"<<endl;
    }
  return 0;
}