HDU 1217 Arbitrage (Floyd)

Arbitrage

http://acm.hdu.edu.cn/showproblem.php?pid=1217

Problem 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

 

 解题思路：将货币种类的名字转化为数字，用其作为货币相互转换的下标，然后用Floyd算出货币之间转换后的最多货币，再寻找自己转换为自己大于1的情况，如果有就输出Yes， 没有就输出No

解题代码：

// File Name: Arbitrage 1217.cpp
// Author: sheng
// Created Time: 2013年07月19日 星期五 15时57分20秒

#include <math.h>
#include <string.h>
#include <stdio.h>
#include <string>
#include <iostream>
#include <map>
using namespace std;

const int max_n = 32;
map<string, int> k;

double cas[max_n][max_n];

int main()
{
    string str1, str2;
    int n, m, T = 1;
    while (scanf("%d", &n) != EOF && n)
    {
        k.clear();
        memset (cas, 0, sizeof (cas));
        for (int i = 1; i <= n; i ++)
        {
            cin >> str1;
            k[str1] = i;//转换为数字
        }
        scanf ("%d", &m);
        while (m --)
        {
            double temp;
            cin >> str1;
            cin >> temp;
            cin >> str2;
            cas[k[str1]][k[str2]] = temp;
        }
        for (int i = 1; i <= n; i ++)
            for (int j = 1; j <= n; j ++)
                for (int k = 1; k <= n; k ++)
                {
                    if (cas[j][k] < cas[j][i] * cas[i][k])
                        cas[j][k] = cas[j][i] * cas[i][k];
                }
        int i;
        for (i = 1; i <= n; i ++)
            if (cas[i][i] > 1)
            {
                printf ("Case %d: Yes
", T++);
                break;
            }
        if (i > n)
            printf ("Case %d: No
", T++);
    }
    return 0;
}