POJ - 2240 Arbitrage

本文介绍了一种通过图论中的Floyd算法检测货币交易是否存在套利机会的方法。该算法利用输入的货币汇率列表构建图,并检查是否存在从一种货币出发经过一系列交易后返回原货币且价值增加的情况。

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Arbitrage
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 26591 Accepted: 11214

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

建图很重要!所有汇率为:自己与自己设置为1,其他就是0。最后套一遍flody把所有汇率兑换的方式遍历一遍,与以往不同

ma[i][j] = ma[i][k]*ma[k][j]

边与边之间应该用乘法来松弛,最后换了一圈回来看看自己与自己的汇率有没有超过1.0

#include<iostream>
#include<cstring>
#include<cstdio>
#include<map>
using namespace std;
int main ()
{

	int n,cut,por = 0;
	while (scanf ("%d",&cut) && cut){
		double ma[35][35] = {0};
		map<string,int> m;
		getchar ();
		for (int i = 1;i <= cut; i++){
			char str[30];
			scanf ("%s",str);
			m[str] = i;
		}
		scanf ("%d",&n);
		getchar ();
		for (int i = 1;i <= n; i++){
			double cur;
			char str[30],str1[30];
			cin >> str >> cur >> str1;
			ma[m[str]][m[str1]] = cur;
		}

		for (int k = 1;k <= cut; k++){
			for (int i = 1;i <= cut; i++){
				for (int j = 1;j <= cut; j++){
					if (ma[i][j] < ma[i][k]*ma[k][j]){
						ma[i][j] = ma[i][k]*ma[k][j];
					}
				}
			}
		}

		for (int i = 1;i <= cut; i++){
			if (ma[i][i] > 1){
				cout << "Case "<< ++por<<": Yes\n";
				break;
			}
			else{
				cout << "Case "<< ++por<<": No\n";
				break;
			}
		}
	}
	return 0;
}

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