ARBITRAG - Arbitrage

no tags 

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. 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
sorablaze_11: 2017-11-21 06:15:51

Very weak test cases.

bhaduriu: 2017-04-19 17:45:17

Bellman-Ford Algorithm, using negative logarithmic edges and then detecting a negative cycle. Done!

vengatesh15: 2017-02-14 20:32:17

AC in 1 go using Floyd warshall..

hackerman97: 2017-01-02 15:58:06

Space after semicolon :| :|

deerishi: 2016-11-02 17:52:25

Awesome Problem! Cool application of Floyd Warshall.

gustavoaca1997: 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

PKJ: 2016-06-20 17:18:27

3
A
B
C

3
A 0.1 B
B 0.2 C
C 50 A

0
This gives Yes for an AC soln.

Shubham Garg: 2016-01-02 20:12:47

@Dimitris - double works

Dimitris Rontogiannis: 2015-09-12 17:34:13

Be careful, double gives WA.
Use long double instead.

Praveen Kulkarni: 2015-06-27 16:46:28

@eightnoteeight my AC solution gives No
Also. I assumed that the graph was strongly connected but still got AC.


Added by:Vincenzo Bonifaci
Date:2011-08-07
Time limit:0.242s
Source limit:50000B
Memory limit:1536MB
Cluster: Cube (Intel G860)
Languages:All except: ASM64
Resource:University of Ulm Local Contest 1996