Arbitrage

                        Arbitrage

 

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 file 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

 

【解法一】
易想到Floyd算法,但时间复杂度为O(n^3)

# include<iostream>
# include<cstdio>
# include <cstring>
# include<map>
# include<string>
using namespace std;
map<string,int>mymap;//首次用map,一次搞定
double Link[100][100];
double Max(double c,double b)
{
    return c>b?c:b;
}
int main()
{
    int nNum,nRate,i,j,k,it1,it2;
    double rate;
    string s1,s2;
    int cnt=1;
    int leap;
    while(cin>>nNum&&nNum)
    {
        leap=0;
        mymap.clear();
        memset(Link,0,sizeof(Link));
        for(i=0;i<nNum;i++)
        {
            cin>>s1;
            mymap[s1]=i;//转化字符串
        }
        cin>>nRate;
        for(i=0;i<nRate;i++)
        {
            cin>>s1>>rate>>s2;
            it1=mymap[s1];
            it2=mymap[s2];
            if( Link[it1][it2]==0)
                Link[it1][it2]=rate;
            else Link[it1][it2]=Max(Link[it1][it2],rate);//连接,赋值
        }
        for(i=0;i<nNum;i++)
        {
            for(j=0;j<nNum;j++)
            {
                if(Link[i][j])
                {
                    for(k=0;k<nNum;k++)
                    {
                        if(Link[j][k])
                        {
                            if(Link[i][k])
                                Link[i][k]=Max(Link[i][k],Link[i][j]*Link[j][k]);//连接,赋值
                            else 
                                Link[i][k]=Link[i][j]*Link[j][k];
                        }
                    }
                }
            }
        }
        for(i=0;i<nNum;i++)
        {
            if(Link[i][i]>1)
            {
                leap=1;
                break;
            }
        }
        if(leap)
            printf("Case %d: Yes\n",cnt++);
        else 
            printf("Case %d: No\n",cnt++);
    }
    return 0;
}

【解法二】
再用Bellman_Ford算法,时间复杂度为O(n^2)

 

# include<iostream>
# include<map>
# include<string>
# define inf 0xffffff
using namespace std;
map<string,int>mymap;
int cnt;
struct node
{
    int start,end;
    double rate;
}num[1000];
double dis[1000];
void bellman_ford(int n,int nNode)
{
    int i,j,flag;
    for(i=0;i<nNode;i++)
    {
        dis[i]=0;
    }
    dis[0]=1;
    for(i=0;i<nNode;i++)//最坏情况遍历的次数
    {
        for(j=0,flag=1;j<n;j++)
        {
            if(dis[num[j].end]<dis[num[j].start]*num[j].rate)//更新
            {
              dis[num[j].end]=dis[num[j].start]*num[j].rate;
              flag=0;
            }
        }
        if(flag)//出现断点
        {
            break;
        }

    }
    for(i=0;i<n;i++)
    {
        if(dis[num[i].end]<dis[num[i].start]*num[i].rate||dis[0]>1)//再次遍历,更新判断
            {
            cout<<"Case "<<cnt++<<": Yes"<<endl;
            return;
            }    
    }
            cout<<"Case "<<cnt++<<": No"<<endl;
}
int main()
{
    string s1,s2;
    int nNum,nRate,i;
    cnt=1;
    while(cin>>nNum,nNum)
    {
        mymap.clear();
        for(i=0;i<nNum;i++)
        {
            cin>>s1;
            mymap[s1]=i;//转化
        }
        cin>>nRate;
        for(i=0;i<nRate;i++)
        {
            cin>>s1>>num[i].rate>>s2;
            num[i].start=mymap[s1];
            num[i].end=mymap[s2];
        }
        bellman_ford(nRate,nNum);
    }
    return 0;
}