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  • poj 2240 floyd算法

    Arbitrage
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 17349   Accepted: 7304

    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
    

    Source

    这一题先要用map将字符转义为数字,每一个字符串用数字代表然后读入图中。

    在这一题求的是从起点開始通过一系列的路径最后回到起点是否能找到一条路使得他们的乘积>1,所以map[i][i]初始化就不是1。而是0.这里的方法是用floyd求出每一个通路的最大值,不断的更新以达到求最大值的目的。

    開始没有想到。想用深搜来实现,但是这样的方法更简单高效

    #include<iostream>
    #include<map>
    #include<string>
    #include<cstdio>
    #include<algorithm>
    #include<cstring>
    using namespace std;
    
    double g[50][50];
    int main()
    {
    	int n,w;
    	double val;
    	string s,s1,s2;
    	map<string,int>q;
    	int cas=0;
    	while(~scanf("%d",&n)&&n)
    	{
    		memset(g,0,sizeof(g));
    		for(int i=0;i<n;i++)
    		{
    			cin>>s;
    			q[s]=i;
    		}
    		scanf("%d",&w);
    		while(w--)
    		{
    			cin>>s1>>val>>s2;
    			g[q[s1]][q[s2]]=val;
    		}
    		
    		for(int k=0;k<n;k++)
    		{
    			for(int i=0;i<n;i++)
    			{
    				for(int j=0;j<n;j++)
    				{
    					g[i][j]=max(g[i][j],g[i][k]*g[k][j]);
    				}
    			}
    		}
    		int ok=0;
    		for(int i=0;i<n;i++)
    		{
    			if(g[i][i]>1)
    				{
    					ok=1;
    					break;
    				}
    		}
    		printf("Case %d: %s
    ",++cas,ok?"Yes":"No");
    	}
    	return 0;
    }

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  • 原文地址:https://www.cnblogs.com/yjbjingcha/p/7000520.html
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