Arbitrage
Time Limit: 2000/1000 MS (Java/others) memory limit: 65536/32768 K (Java/Others)
Total submission (s): 3277 accepted submission (s ): 1489 problem descriptionarbitrage 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 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, 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.
Inputthe input file will contain in one or more test cases. om 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.
Outputfor each test case, print one line telling whether Arbitrage is possible or not in the format "case: yes" respectively "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
Sourceuniversity of Ulm Local contest 1996
Idea: bellmanford used to determine whether there is a negative weight loop. Now, it is only enough to find whether there is a positive weight loop. Code:
#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <map>#define maxn 35using namespace std;int n,m,ans;double r;double dist[maxn];double city[maxn][maxn];string s,s1,s2;map<string,int>money;struct Node{ int le,ri; double rate;}edge[maxn*maxn];void init(){ money.clear(); memset(city,0,sizeof(city)); memset(dist,0,sizeof(dist)); dist[1]=1;}bool BellmanFord(){ int i,j; for(i=1;i<n;i++) { for(j=1;j<=m;j++) { if(dist[edge[j].ri]<dist[edge[j].le]*edge[j].rate) dist[edge[j].ri]=dist[edge[j].le]*edge[j].rate; } } for(j=1;j<=m;j++) { if(dist[edge[j].ri]<dist[edge[j].le]*edge[j].rate) return true ; } return false ;}int main(){ int i,j,t=0; while(scanf("%d",&n),n) { t++; init(); for(i=1;i<=n;i++) { cin>>s; money[s]=i; } scanf("%d",&m); for(i=1;i<=m;i++) { cin>>s1>>r>>s2; edge[i].le=money[s1]; edge[i].ri=money[s2]; edge[i].rate=r; } printf("Case %d: ",t); if(BellmanFord()) printf("Yes\n"); else printf("No\n"); } return 0;}