Arbitrage
Time limit:1000ms Memory limit:65536k
Total submissions:16429 accepted:6909
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
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
ULM Local 1996
#include <iostream> #include <cstdio> #include <cstdlib> #include <algorithm> #include <map > #include <string>using namespace std;double cost[32][32];d ouble dis[32];int n,m;map<string,int> Mapstring;bool Bellman_ford () {for (Int. k=1;k<=n;k++) for (int. i=1;i<=n;i++) for (int j=1;j<=n;j + +) {if (Cost[i][j]<cost[i][k]*cost[k][j]) cost[i][j]=cost[i][k]*cost[k][j]; } for (int i=1;i<=n;i++) if (cost[i][i]>1) return true; return false;} int main () {//Freopen ("In.txt", "R", stdin); String s,s1,s2; Double C;int cnt=1; while (1) {scanf ("%d", &n); if (n==0) break; for (int i=1;i<=n;i++) {cost[i][i]=1; cin>>s; Mapstring[s]=i; } scanf ("%d", &m); for (int i=1;i<=m;i++) {cin>>s1>>c>>s2; Cost[mapstring[s1]][mapstring[s2]]=c; } int Ans=bellman_foRD (); if (ans) printf ("Case%d:yes\n", cnt++); else printf ("Case%d:no\n", cnt++); } return 0;}
POJ 2240 Bellman_ford