poj 2240 Arbitrage

Arbitrage Time Limit:1000MS    Memory Limit:65536KB    64bit IO Format:%I64d & %I64u

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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
题目意思是给你n种货币的名称，然后在给你m种货币之间的转换。问你是否存在一种货币经过转换后
价值变高了。就是刚开始自身价值都为1，转化了一圈后价值会不会大于1。如果存在则输出Yes，否则输出No。

思路：这里因为数据不是很大，所以直接用flody算法好了。用之前先用map容器把货币用数字来表示一下。
     
     
     
     

      
      
      
      
 
       
 
        
Memory: 228 KB
 
        
 
 
        
Time: 79 MS
 
       
 
      
     
     
     
     
#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include <map>
#include <string>
#include <string.h>
using namespace std;
map<string,int>mp;
const int nmax = 33;
const int inf =0xfffffff;
double temp[nmax][nmax];
int top;
void flody(int n)
{
		int i,j,k;
		for(i=1;i<=n;i++)
		{
			for(j=1;j<=n;j++)
			{
				for(k=1;k<=n;k++)
				{
					if(temp[i][j]<temp[i][k]*temp[k][j])
							temp[i][j]=temp[i][k]*temp[k][j];	
				}
			}
		}	
}
int main()
{
	int n,i,m,j;
	string str,a,b;
	double val;
	int qqq=1;
	while(scanf("%d",&n)!=-1&&n)
	{
	
		int tt=1;
	
		for(i=0;i<n;i++)
		{		
			cin>>str;
			mp[str]=tt++;
	
		}
		scanf("%d",&m);
		top=0;
		for(i=1;i<tt;i++)
			for(j=1;j<tt;j++)
			{
				if(i==j)
					temp[i][j]=1;//自身价值都为1.
				else
					temp[i][j]=0;//因为要找最大值，所以初始化为0而不是无穷。
			}
		while(m--)
		{
			cin>>a>>val>>b;
				temp[mp[a]][mp[b]]=val;
		}
		flody(tt-1);
		int ans=0;
		for(i=1;i<tt;i++)
		{
				if(temp[i][i]>1)
				{
					ans=1;
					break;
				}
		}
		if(ans)
			printf("Case %d: Yes\n",qqq++);
		else
			printf("Case %d: No\n",qqq++);	
	}
return 0;	
}