1008 最短路

最新推荐文章于 2022-05-11 21:56:05 发布

xiaoleiacm

最新推荐文章于 2022-05-11 21:56:05 发布

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分类专栏：弗洛伊德

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本文介绍了一种通过检测不同货币间的兑换率是否存在盈利机会的方法。利用 Floyd 算法进行两两货币之间的比较，判断是否能通过一系列转换使得初始货币金额增加。

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Arbitrage

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 7 Accepted Submission(s) : 5

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：最短路么~ 就是看看一个货币经过兑换之后能不能变成比原来的更多

用floyd进行两两之间的比较

重点：map的应用

代码：

#include<iostream>
#include<map>
#include<string>
using namespace std;
const int maxn=10000;
double map1[maxn][maxn];
int n;
void floyd()//只能 用floyd 因为最后是相乘的形式~ 对于1来说已经是算是负值了 
{
	int k,i,j;
	for(k=1;k<=n;k++)
	 for(i=1;i<=n;i++)
	  for(j=1;j<=n;j++)
	   {
   		if(map1[i][j]<map1[i][k]*map1[k][j])
   		  map1[i][j]=map1[i][k]*map1[k][j];   		 
   	   }
}
int main()
{
   	map<string,int> mp; //map映射 省去了一个下标函数 
	char str[1000];
	char str1[1000],str2[1000];
	double huilv;
	int n2;
	int flag;
	int t=0;
     while(cin>>n,n)
     {
     	 mp.clear();
     	 
    	 for(int i=1;i<=n;i++)
    	  for(int j=1;j<=n;j++)
    	  {
  	    	map1[i][j]=1.0;
  	    }
     	 for(int j=1;j<=n;j++) 
   	     {
   	    	cin>>str;
   	    	mp[str]=j;
   	    //	cout<<mp[str]<<endl;
     	 }
     	 cin>>n2;
    	 for(int i=1;i<=n2;i++)
    	 {
    	 	cin>>str1>>huilv>>str2;
    	 
    	 	int a=mp[str1];
    	 	int b=mp[str2];
    		    map1[a][b]=huilv;
	//	cout<<map1[mp[str1]][mp[str2]]<<endl;	  	    	
 	     }
 	    
         
        
        /*
for(int i=1;i<=n;i++)
        {     	 
 	       for(int j=1;j<=n;j++)
 	       {
       	     	cout<<map1[i][j]<<" ";
       	   }
       	     cout<<endl;
 	     }*/
		  flag=0;
    	floyd();
    	for(int i=1;i<=n;i++)
    	{
	    	if(map1[i][i]>1.0)
	    	  {
  	    		  flag=1;
	    	break;
  	    	}
	    }
	    
	    
	   if(flag)
	  cout<<"Case "<<++t<<": Yes"<<endl; 
		else
		   cout<<"Case "<<++t<<": No"<<endl;
     } 
	 
     return 0;
	
}