ARBITRAG  Arbitrage
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 pounds, 1 British pound buys 10.0 French francs, and 1 French franc buys 0.21 US dollars. 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. On 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 c_{i} of a source currency, a real number r_{ij} which represents the exchange rate from c_{i} to c_{j} and a name c_{j} of the destination currency. Note that c_{i} and c_{j} may be the same 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".
Example
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 Output: Case 1: Yes Case 2: No
hide comments
shohan2032:
20231010 14:30:26
can any body give me the bellman ford solution for this problem??Though i have solved it with warshall. 

ive1010:
20220322 16:34:12
Bellman ford with time complexity : O(N^4) 

vineetjai:
20200724 14:37:20
Don't think of taking log or finding cycles and don't try to convert it into bellman ford. Use Floyd Warshall for a quick answer. 

roopammishra:
20200323 22:55:54
Wait ! What !..


enigmus:
20190606 10:56:26
Printing uppercase answers costed me a couple of WAs :(.


abhinav_jain02:
20190523 14:28:31
AC in one go


surajkumar_cse:
20190115 23:13:02
The problem demands to product the weight of the edges ,,, after lots of thinking and online help ... I was able to convert the product into sum. Hint: 12th standard maths. 

jmr99:
20180815 15:23:05
FloyWar!! 

abdelhameedddd:
20180807 10:35:12
AC USING FLOYD :) 

anirudnits:
20180618 15:04:47
@siva2697 thanks and my first AC using Floyd warshall :) 
Added by:  Vincenzo Bonifaci 
Date:  20110807 
Time limit:  1s 
Source limit:  50000B 
Memory limit:  1536MB 
Cluster:  Cube (Intel G860) 
Languages:  All except: ASM64 
Resource:  University of Ulm Local Contest 1996 