Algorithm
Problem link- https://www.spoj.com/problems/ARBITRAG/
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
Code Examples
#1 Code Example with C++ Programming
Code -
C++ Programming
#include <iostream>
#include <queue>
#include <vector>
#include <map>
using namespace std;
int N,M;
map<string, int> nodes;
double v[300][300];
int main ()
{
string a,b,c;
int i,j;
double k;
int cas=1;
while(1){
nodes.clear();
cin >> N;
if(N==0)
break;
for(i=0;i<N;i++){
cin >> a;
nodes[a]=i;
}
for(i=0;i<N;i++)
for(j=0;j<N;j++){
v[i][j]=0.0;
}
cin >> M;
for(int z=0;z<M;z++){
cin >> a >> k >> b;
i=nodes.find(a)->second;
j=nodes.find(b)->second;
v[i][j]=k;
}
for(int via=0; via<N;via++)
for(int f=0;f<N;f++)
for(int t=0;t<N;t++)
if(v[f][t] < v[f][via]*v[via][t])
v[f][t] = v[f][via]*v[via][t];
for(i=0;i<N;i++)
if(v[i][i]>1.0)
break;
cout << "Case " << cas << ": ";
if(i==N)
cout << "No" << endl;
else
cout << "Yes" << endl;
cas++;
}
return 0;
}Input
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 2: No
#2 Code Example with Java Programming
Code -
Java Programming
/*
* To change this template, choose Tools | Templates
* and open the template in the editor.
*/
import java.io.BufferedReader;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.util.HashMap;
import java.util.Map;
import java.util.StringTokenizer;
/**
*
* @author mdshohag
*/
public class Main{
public static void main(String[] args) {
InputStream stream=System.in;
OutputStream outputstream=System.out;
InputReader in=new InputReader(stream);
int test=0;
while (true) {
int n=in.NextInt();
if(n==0)break;
String s;
Map<String,Integer>mymap=new HashMap<String, Integer>();
mymap.clear();
for(int i=0;i<n;i++)
{
s=in.Next();
mymap.put(s, i);
}
int m;
String source,distantion;
double cost;
double [][]data=new double[50][50];
for(int i=0;i<n;i++)
for(int j=0;j<n;j++)
data[i][j]=0.0;
m=in.NextInt();
for(int i=0;i<m;i++)
{
source=in.Next();
cost=in.NextDouble();
distantion=in.Next();
int x,y;
x=mymap.get(source);
y=mymap.get(distantion);
data[x][y]=cost;
}
for(int k=0;k<n;k++)
for(int i=0;i<n;i++)
for(int j=0;j<n;j++)
if(data[i][j]<data[i][k]*data[k][j])
data[i][j]=data[i][k]*data[k][j];
boolean flag=false;
for(int i=0;i<n;i++)
if(data[i][i]>1.0)
{
flag=true;
break;
}
if(flag)
{
System.out.printf("Case %d: Yes\n", ++test);
}
else
{
System.out.printf("Case %d: No\n", ++test);
}
}
}
}
class InputReader
{
public BufferedReader reader;
public StringTokenizer tokenizer;
public InputReader(InputStream stream)
{
reader=new BufferedReader(new InputStreamReader(stream));
tokenizer=null;
}
public String Next()
{
while(tokenizer==null || !tokenizer.hasMoreTokens())
{
try {
tokenizer=new StringTokenizer(reader.readLine());
} catch (IOException e) {
throw new RuntimeException();
}
}
return tokenizer.nextToken();
}
public int NextInt()
{
return Integer.parseInt(Next());
}
public double NextDouble()
{
return Double.parseDouble(Next());
}
public float NextFloat()
{
return Float.parseFloat(Next());
}
}Input
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 2: No
Demonstration
SPOJ Solution-Arbitrage-Solution in C, C++, Java, Python