Arbitrage

Time Limit: 1000MS	Memory Limit: 65536K
Total Submissions: 12761	Accepted: 5392

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

題目大意及其思路：問你能不能找到一個圈，使沿著這個圈兌換錢幣後錢數增加。

解法：路徑為匯率，由於自環匯率是1，因此，我們要找一個圈作用的結果大於1。因為題目給得很清楚，是邊邊相乘，於是就是路徑鬆弛的時候改成相乘就可以了。下面我用map進行字串與節點編號的對映。（一開始習慣性的把G[][]宣告int.....好腦殘啊）

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <ctime>
#include <queue>
#include <stack>
#include <vector>
#include <algorithm>
#include <string>
#include <map>
using namespace std;
const int maxn=40;
const int inf=-10000000;
double G[maxn][maxn];
int n,m;
map<string, int > Map;    //終於知道STL的強大之處了，雖然慢了一點
int main()
{
    freopen("in.txt","r",stdin);
	string s,sa,sb;
	int cas=0;
	double rate;
	while(scanf("%d",&n)!=EOF&&n!=0)
	{
		memset(G,0,sizeof(G));
		Map.clear();
		for(int i=1;i<=n;i++)
		{
			for(int j=1;j<=n;j++)
			{
				if(i==j)	G[i][j]=1;
				else	G[i][j]=-inf;
			}
		}
		for(int i=1;i<=n;i++)
		{
			cin>>s;
			Map[s]=i;     //map的對映關係
		}
		cin>>m;
		for(int i=1;i<=m;i++)
		{
			cin>>sa>>rate>>sb;
			G[Map[sa]][Map[sb]]=rate;
			//cout<<Map[sa]<<' '<<Map[sb]<<' '<<rate<<endl;
			//cout<<G[Map[sa]][Map[sb]]<<endl;
		}
		/*                         //開關除錯法，輸出中間值，很有用的哦
		for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
                cout<<i<<' '<<j<<' '<<G[i][j]<<endl;
        }
        //*/
		for(int k=1;k<=n;k++)         //標準的Floyd
		{
			for(int i=1;i<=n;i++)
			{
				for(int j=1;j<=n;j++)
				     if(G[i][j]!=inf&&G[i][k]!=inf&&G[k][j]!=inf)
                                              G[i][j]=max(G[i][k]*G[k][j],G[i][j]);
			}
		}

		/*cout<<endl;
		for(int i=1;i<=n;i++)
                {
                 for(int j=1;j<=n;j++)
                        cout<<i<<' '<<j<<' '<<G[i][j]<<endl;
                 }//*/ 

		bool is_ok=false;
		for(int i=1;i<=n;i++)
		{
		    if(G[i][i]>1)
                     {
                          is_ok=1;
                          break;
                     }         
		}
		if(is_ok)	printf("Case %d: Yes\n", ++cas);
		else 	printf("Case %d: No\n", ++cas);
	}
    return 0;
}