hdu 1217 Arbitrage

最新推荐文章于 2020-11-10 19:47:30 发布

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Arbitrage

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8579 Accepted Submission(s): 3956

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


预备知识:stl中的map<string,int>xxx(表示建立一个以字符串为下标，以十进制整数为值的图)
方法:根据回了建立一张图，根据题目找出最大环(权值相乘),若权值大于1就可以有收益
#include<iostream>
#include<cstdio>
#include<map>
#include<cstring>
using namespace std;
double mp1[100][100];
int main(int argc, char *argv[])
{
	int i,n,j;
	int a[10000];
	char b[100];
	char c[100];
	map<string,int>mymap;
	int flag1=1;
	while(scanf("%d",&n),n)
	{
		for(i=0;i<n;i++)
		{
			scanf("%s",b);
			mymap[b]=i;
		}
		for(i=0;i<n;i++)
		{
			for(j=0;j<n;j++)
			{
				mp1[i][j]=0;
			}
		}
		int m;
		double h;
		scanf("%d",&m);
		getchar();
		for(i=0;i<m;i++)
		{
			scanf("%s %lf %s",b,&h,c);
			mp1[mymap[b]][mymap[c]]=h;	
		}
		int k;
		for(k=0;k<n;k++)
		{
			for(i=0;i<n;i++)
			{
				for(j=0;j<n;j++)
				{
					mp1[i][j]=max(mp1[i][j],mp1[i][k]*mp1[k][j]);
				}
			}
		}
		int flag=0;
		for(i=0;i<n;i++)
		{
			if(mp1[i][i]>1)
			{
				flag=1;
			}
		}
		printf("Case %d: ",flag1++);
		if(flag==1)
		printf("Yes\n");
		else
		printf("No\n");
	}
	return 0;
}