poj2240

Arbitrage

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 18810 Accepted: 7962

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

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

Source

题意：

货币之间可以彼此兑换，问是否有这样一种货币X，经过一些列兑换后，最终得到X*比原来的X多，（兑换了一圈，钱变多了）

由于n<=30,可以用floyd算法较简单，构造邻接矩阵，对角线上值如果有大于一的说明存在这种货币;

用到map使得代码更简洁

#include <iostream>#include <map>#include <string>#include <string.h>using namespace std;double map1[50][50];map<string,int>L;int main(){    int n,m;    string s,s1,s2;    double num;    int tot=1;    while(cin>>n)    {        if(n==0)            break;        for(int i=1; i<=n; i++)        {            cin>>s;            L[s]=i;            map1[i][i]=1;        }        cin>>m;        while(m--)        {            cin>>s1>>num>>s2;            map1[L[s1]][L[s2]]=num;        }        for(int i=1; i<=n; i++)        {            for(int j=1; j<=n; j++)            {                for(int k=1; k<=n; k++)                {                    if(map1[j][k]<map1[j][i]*map1[i][k])                        map1[j][k]=map1[j][i]*map1[i][k];                }            }        }        int flag=0;        for(int i=1; i<=n; i++)        {            //cout<<map1[i][i]<<" ";            if(map1[i][i]>1.0)            {                flag=1;            }        }        //  cout<<endl;        cout<<"Case "<<tot++<<": ";        if(flag==1)            cout<<"Yes"<<endl;        else            cout<<"No"<<endl;    }    return 0;}