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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 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. Note that ci and cj 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
| Added by: | Vincenzo Bonifaci |
| Date: | 2011-08-07 |
| Time limit: | 1s |
| Source limit: | 50000B |
| Memory limit: | 1536MB |
| Cluster: | Cube (Intel G860) |
| Languages: | All except: ASM64 |
| Resource: | University of Ulm Local Contest 1996 |
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2018-03-30 13:37:37
I see small constraint I use Floyd Warshall :P. AC in 1 go :D |
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2018-03-17 06:57:45
spoj tool kit wont be helpful after reading edges dont intialize dist[i][i]=1 bcoz input contains self loop. Last edit: 2018-03-17 07:02:31 |
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2018-03-17 06:57:03
incase if u assume that dollar to dollar intial is 1 it isnt working dont intialize dist[i][i]=1; and dont use spoj toolkit for this problem giving wrong results. |
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2017-11-21 06:15:51
Very weak test cases. |
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2017-04-19 17:45:17
Bellman-Ford Algorithm, using negative logarithmic edges and then detecting a negative cycle. Done! |
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2017-02-14 20:32:17
AC in 1 go using Floyd warshall.. |
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2017-01-02 15:58:06
Space after semicolon :| :| |
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2016-11-02 17:52:25 deerishi
Awesome Problem! Cool application of Floyd Warshall. |
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2016-09-01 10:00:37
input: 4 A B C D 5 A 0.5 B B 0.5 C C 1 D D 2 A A 0.9 C 0 Output: Case 1: Yes |
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2016-06-20 17:18:27 PKJ
3 A B C 3 A 0.1 B B 0.2 C C 50 A 0 This gives Yes for an AC soln. |
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