"POJ 2240" Arbitrage

Arbitrage

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 18408		Accepted: 7796

Description

Arbitrage is the use of discrepancies in currency exchange rates to transform one unit of a currency into more than one UN It's 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, MA King a profit of 5 percent.

Your job is to write a program this takes a list of currency exchange rates as input and then determines whether arbitrage is possible or not.

Input

The input would contain one or more test cases. Om the first line of all test case there was an integer n (1<=n<=30), representing the number of different Currencie S. The next n lines each contain the name of one currency. Within a name no spaces would appear. The next line contains one integer m, representing the length of the table to follow. The last m lines all 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 is impossible.
Test cases is separated from 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 are possible or not in the format "case Case:yes" Respectivel Y "Case Case:no".

Sample Input

3usdollarbritishpoundfrenchfranc3usdollar 0.5 britishpoundbritishpound 10.0 Frenchfrancfrenchfranc 0.21 Usdollar3usdollarbritishpoundfrenchfranc6usdollar 0.5 britishpoundusdollar 4.9 frenchfrancbritishpound 10.0 Frenchfrancbritishpound 1.99 usdollarfrenchfranc 0.09 Britishpoundfrenchfranc 0.19 USDollar0

Sample Output

Case 1:yescase 2:no

Source

ULM Local 1996 can be used Floyd. It is important to note that he is seeking a longest path like the ring. So this is different from the normal Floyd, but it's essentially a transitive closure.

#include <cstdio>#include<cstring>#include<iostream>#include<string>#include<map>using namespaceStd;map<string,int>ID;Doubledis[ *][ *];intN, M;intMain () {intKase =0;  while(SCANF ("%d", &n) = =1&&N) {strings, S1;        Id.clear ();  for(inti =0; I < n; ++i) {cin>>s; Id[s]=i; }         for(inti =0; I < n; ++i) for(intj =0; J < N; ++j) Dis[i][j]=1.0; scanf ("%d", &m);  for(inti =0; I < m; ++i) {Doublex; CIN>> s >> x >>S1; intID1 = Id[s], Id2 =ID[S1]; DIS[ID1][ID2]=x; }         for(intK =0; K < n; ++k) for(inti =0; I < n; ++i) for(intj =0; J < N; ++j) Dis[i][j]= Max (Dis[i][j], Dis[i][k] *Dis[k][j]); BOOLFlag =true; printf ("Case %d:", ++Kase);  for(inti =0; I < n; ++i)if(Dis[i][i] >1) {flag=false; printf ("yes\n");  Break; }        if(flag) printf ("no\n"); }    return 0;}

"POJ 2240" Arbitrage