POJ

题意：

通过转换货币，一个聪明的交易者可以从1美元开始，买入0.5×10×0.21＝1.05美元，获利5%。问转换货币从美元到美元是否能够获利。

输入中有字符串，可以用Map<string,int>p,处理成数字，用floyd()求解两者转换最大的。

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

3USDollarBritishPoundFrenchFranc3USDollar 0.5 BritishPoundBritishPound 10.0 FrenchFrancFrenchFranc 0.21 USDollar3USDollarBritishPoundFrenchFranc6USDollar 0.5 BritishPoundUSDollar 4.9 FrenchFrancBritishPound 10.0 FrenchFrancBritishPound 1.99 USDollarFrenchFranc 0.09 BritishPoundFrenchFranc 0.19 USDollar0

Sample Output

Case 1: YesCase 2: No

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<map>#define MAXV 35#define inf 1<<30using namespace std;double Map[MAXV][MAXV];map<string,int>p;int n;void floyd(){    for(int k=1;k<=n;k++)    {        for(int i=1;i<=n;i++)        {            for(int j=1;j<=n;j++)            {                if(Map[i][j]<Map[i][k]*Map[k][j])                    Map[i][j]=Map[i][k]*Map[k][j];            }        }    }}int main(){    int time=1;    int m;    string s;    while(cin>>n,n)    {        for(int i=1;i<=n;i++)        {            cin>>s;            p[s]=i;            Map[i][i]=1;        }        string s1,s2;        double money;        cin>>m;        for(int i=1;i<=m;i++)        {            cin>>s1>>money>>s2;            Map[p[s1]][p[s2]]=money;        }        floyd();        int flag=0;        cout<<"Case "<<time++<<": ";        for(int i=1;i<=n;i++)        {            if(Map[i][i]>1)            {                cout<<"Yes"<<endl;                flag=1;                break;            }        }        if(!flag)            cout<<"No"<<endl;    }    return 0;}