POJ-2240 Arbitrage(Floyd)

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本文介绍了一种利用Floyd算法变形来检测是否存在货币兑换仲裁机会的方法。通过输入不同货币之间的兑换率，程序能够判断是否可以通过特定的货币转换路径实现盈利。

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原题：

Arbitrage

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 20551		Accepted: 8750

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 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

题意：

给出货币A到货币B的汇率，求能不能通过汇率转换使自己的货币增多。

思路：

Floyd算法的变形，把判断条件改成了乘法就可以了（dis[i][j] < dis[i][k]*dis[k][j]）。

代码：

#include <iostream>
#include <map>
#include <string.h>
using namespace std;
const int inf = 0x3f3f3f3f;
int n;
double dis[31][31];
map<string,int>cur;

void floyd()
{
	for(int k = 0;k < n;k++)
	for(int i = 0;i < n;i++)
	{
		for(int j = 0;j < n;j++)
		{
			if(dis[i][j] < dis[i][k]*dis[k][j])
				dis[i][j] = dis[i][k]*dis[k][j];
		}
	}
}

int main()
{
    while(cin>>n,n)
	{
		static int count = 1;
		memset(dis,0,sizeof(dis));
		string t;
		for(int i = 0;i < n;i++)
		{
			cin>>t;
			cur[t] = i;
			dis[i][i] = 1;
		}
		int m;
		cin>>m;
		while(m--)
		{
			string a, b;
			double rate;
			cin>>a>>rate>>b;
			dis[cur[a]][cur[b]] = rate;
		}
		floyd();
		bool flag = false;
		for(int i = 0;i < n;i++)
			if(dis[i][i]>1)
				{flag = true;break;}
		if(flag)
			cout<<"Case "<<count++<<": "<<"Yes"<<endl;
		else
			cout<<"Case "<<count++<<": "<<"No"<<endl;
	}
    return 0;
}