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

Ulm Local 1996

这一题先要用map将字符转义为数字，每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开始通过一系列的路径最后回到起点能否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求出每个通路的最大值，不断的更新以达到求最大值的目的。开始没有想到，想用深搜来实现，可是这种方法更简单高效

#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\n",++cas,ok?"Yes":"No");
	}
	return 0;
}