集訓隊專題（3）1001 Arbitrage

Arbitrage

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 6037    Accepted Submission(s): 2797

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​​

      此題把匯率看做路程就可以将此題轉換成最短(長)路徑問題，由于此題需要知道每一種貨币和自己的最大轉換，是以這裡我們最好用Floyd算法，當然能用這種算法還有一個最大的原因是此題的資料比較弱，Floyd算法的複雜度為O(n^3)，而此處的n最大隻有30（當然Floyd算法的核心算法代碼隻有5行，寫起來如此友善，當然優先選這個^v^~）

#include <cstdio>
#include <cstring>
#include <string>
#include <map>
#include <iostream>
#include <algorithm>
using namespace std;
map<string, int> Map;
const int maxn = 35;
double G[maxn][maxn];
string str;
int n;
void Floyd()
{
  for(int i=1; i<=n; i++)
    for(int j=1; j<=n; j++)
      for(int k=1; k<=n; k++)
        if(G[j][k] < G[j][i]*G[i][k])
          G[j][k] = G[j][i]*G[i][k];
}
int main()
{
  int t=1;
  while(scanf("%d",&n) && n)
  {
    memset(G,0,sizeof(G));
    Map.clear();
    for(int i=1; i<=n; i++)
    {
      cin >> str;
      Map[str] = i;
    }
    int m;
    scanf("%d",&m);
    while(m--)
    {
      string a,b;
      double rate;
      cin >> a >> rate >> b;
      int x = Map[a];
      int y = Map[b];
      G[x][y] = rate;
    }
    Floyd();
    bool ok = 0;
    for(int i=1; i<=n; i++)
      if(G[i][i] > 1)
      {
        ok = 1;
        break;
      }
    if(ok) printf("Case %d: Yes\n",t++);
    else printf("Case %d: No\n",t++);
  }
}