1224: Arbitrage (II)

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本文介绍了一种通过图算法检测货币兑换中是否存在套利机会的方法。该算法接收一组货币及其兑换率作为输入，并利用Floyd算法寻找是否存在通过一系列兑换使得初始货币数量增加的可能性。

1224: Arbitrage (II)

Result	TIME Limit	MEMORY Limit	Run Times	AC Times	JUDGE
	3s	8192K	265	97	Standard

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 Specification

The input file will contain one or more test cases. On the first line of each test case there is an integer n ( ), 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 of a source currency, a real number which represents the exchange rate from to and a name 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 Specification

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

#include<iostream>
#include<map>
using namespace std;
int main()
{
    int n,m,l=1;
    while(scanf("%d",&n)==1&&n)
    {
        double a[50][50]={0};//floyd 求最长路
        map<string,int> q;
        for(int i=0;i<n;i++)
        {
            string str;
            cin>>str;
            //q.insert(map<string,int>::value_type(str,i));
           // q.insert(pair<string,int>(str,i));
           q[str]=i;
        }
        scanf("%d",&m);
        while(m--)
        {
            string str1,str2;
            double num;
            cin>>str1>>num>>str2;
            a[q[str1]][q[str2]]=num;
        }
        for(int k=0;k<n;k++)
            for(int i=0;i<n;i++)
                for(int j=0;j<n;j++)
                    if(a[i][j]<a[i][k]*a[k][j]) a[i][j]=a[i][k]*a[k][j];
        int flag=1;
        for(int i=0;i<n;i++)
        {
            if(a[i][i]>1) {flag=0;break;}
        }
        if(!flag) printf("Case %d: Yes/n",l++);
        else printf("Case %d: No/n",l++);
    }
    return 0;
}