POJ 2240 Arbitrage

Arbitrage

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 16457		Accepted: 6926

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

题意：给出几种货币以及它们兑换时的汇率，问是否可以经过兑换达到盈利

#include <iostream>
#include <stdio.h>
#include <math.h>
#include <string.h>

using namespace std;

int n,m;
double rate;
int s,t;
char str1[100],str2[100];
char cur[50][100];
double dis[50][50];

void Floyd()
{
    for(int k=1; k<=n; k++)
    {
        for(int i=1; i<=n; i++)
        {
            for(int j=1; j<=n; j++)
            {
                if(dis[i][j]<dis[i][k]*dis[k][j])
                    dis[i][j]=dis[i][k]*dis[k][j];
            }
        }
    }
}

int main()
{
    int sum=0;
    while(scanf("%d",&n)!=EOF)
    {
        if(!n)
            break;
        memset(dis,0,sizeof(dis));
        for(int i=1; i<=n; i++)
        {
            for(int j=1; j<=n; j++)
            {
                if(i==j)
                    dis[i][j]=1;
            }
        }
        for(int i=1; i<=n; i++)
        {
            scanf("%s",cur[i]);
        }
        scanf("%d",&m);
        while(m--)
        {
            scanf("%s%lf%s",str1,&rate,str2);
            int ans=0;
            for(int i=1; i<=n; i++)
            {
                if(strcmp(str1,cur[i])==0)
                {
                    s=i;//找出这是第几种货币，方便下面对货币汇率的记录
                    ans++;
                }
                if(strcmp(str2,cur[i])==0)
                {
                    t=i;
                    ans++;
                }
                if(ans==2)
                    break;
            }
            dis[s][t]=rate;
        }
        Floyd();
        int flag=0;
        for(int i=1; i<=n; i++)
        {
            if(dis[i][i]>1)//如果一种货币到他本身的距离大于1，说明可以盈利
            {
                flag=1;
                break;
            }
        }
        if(flag)
            printf("Case %d: Yes\n",++sum);
        else
            printf("Case %d: No\n",++sum);
    }
    return 0;
}