HDOJ 1217 Arbitrage(拟最短路，floyd算法)

Posted 2020-09-28 yxysuanfa

Arbitrage

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5679    Accepted Submission(s): 2630

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

 

 

 

题意:      套汇是指利用不同外汇市场的外汇差价。在某一外汇市场上买进某种货币，同一时候在还有一外汇市场上卖出该种货币。以赚取利润。这样的利润称之为套利。

比方1美元能够买0.5英镑。而1英镑能够买10法郎，1法郎能够买0.21美元那么可用通过套汇使用1美元买到1.05美元，套利是存在的。以下给出各个货币的种类和名称，再给出一些货币兑换的汇率。请问是否存在套利？

 

解题思路：能够看做最短路问题，只是权值的计算是乘而不是相加。而与小于1的权值相乘会导致权值减小，即存在负权值问题。

 

所以此题不能用dijkstra算法，此题能够用floyd算法。

 

 

代码例如以下：

 

<span style="font-size:12px;">#include<cstdio>
#include<cstring>
#define maxn 32
double map[maxn][maxn];
int n,t;

void floyd()
{
	int i,j,k;
	for(k=0;k<n;++k)
	{
		for(i=0;i<n;++i)
		{
			for(j=0;j<n;++j)
			{
				if(map[i][j]<map[i][k]*map[k][j])
				   map[i][j]=map[i][k]*map[k][j];
			}
		}
	}
	int sign=0;
	for(i=0;i<n;++i)
	{
		if(map[i][i]>1)//自身的汇率为1。若汇率大于1则说明能套汇 
		{
			sign=1;
			break;
		}  
	}
	if(sign)
	   printf("Yes\n");
	else 
	   printf("No\n");
}

int main()
{
	int m,i,j,a,b,t=1;;
	double c;
	char str[32][32],s1[32],s2[32];
	while(scanf("%d",&n)&&n)
	{
		for(i=0;i<n;++i)
		{
			for(j=0;j<n;++j)
				map[i][j]=0;
		}
		for(i=0;i<n;++i)
			scanf("%s",str[i]);
		scanf("%d",&m);
		while(m--)
		{
			scanf("%s%lf%s",s1,&c,s2);
			for(i=0;i<n;++i)
			{
				if(strcmp(s1,str[i])==0)
					a=i;
				if(strcmp(s2,str[i])==0)
					b=i;
			}
			map[a][b]=c;
		}
		printf("Case %d: ",t++);
		floyd();
	}
	return 0;
}</span>