Arbitrage+floyd算法

最新推荐文章于 2023-06-25 12:01:09 发布

李逝水

最新推荐文章于 2023-06-25 12:01:09 发布

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分类专栏：最短路径文章标签： floyd算法 algorithm

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本文介绍了一种通过图算法检测货币兑换中是否存在套利机会的方法。利用弗洛伊德算法计算不同货币之间的转换率，判断是否存在闭环使得起始货币经过一系列转换后能够以大于初始值的金额返回，以此来确定是否有机会进行套利。

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Arbitrage

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3019 Accepted Submission(s): 1365

Problem Description

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 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.

Input

The input file will contain 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.

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".

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

#include<iostream>
#include<map>
#include<string>
using namespace std;
#define MAX 35
int n ;
double s[MAX][MAX];
void floyd()
{
	int i, j, k ;
	for( i = 1 ; i<= n  ; i++)
		for( j = 1 ; j<= n ; j++)
				for( k = 1; k<= n ; k++)
					if((s[j][i]*s[i][k]) > s[j][k])
						s[j][k] = s[j][i] * s[i][k] ;
	for( i = 1; i<= n ; i++ )
		if(s[i][i] > 1.0 )
		{
				cout<<"Yes"<<endl;
				break;
		}
		if(i>n)
			cout<<"No"<<endl;
}
int main(void)
{
	int count = 1;
	while(cin>>n && n)
	{
		int  i , j , l = 1;
		map<string,int>m ;
		char A[102] ,B[102];
		double p;
		for( i = 0 ;i<MAX ; i++)
			for( j = 0 ; j< MAX ; j++) 
				if(i==j)
					s[i][j] = 1;
				else
					s[i][j]  = 0;
		for( i = 0 ; i< n ; i++)
		{
			cin>>A;
			if(!m[A])	
				m[A] = l++;
		}
		int m1 ;
		cin >> m1 ;
		for( i = 0 ; i < m1 ; i++)
		{
			cin>>A >> p >> B ;
			s[m[A]][m[B]] = p;
		}
		cout<<"Case "<<count ++ <<": ";
		floyd();
	}
	return 0;
}