hdu 1217 Arbitrage

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

Arbitrage
Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
Submit Status

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

这是一个货币交换问题，就是问你货币交换之后能不能赚到钱，用num函数将下标由字符串转换为数字，g

表示两种货币之间的汇率

参考别人的代码

<pre name="code" class="cpp">#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <limits>
#include <queue>
#include <stack>
#include <vector>
#include <map>

using namespace std;

#define N 1100
#define INF 0x3f3f3f3f
#define PI acos (-1.0)
#define EPS 1e-8

int n, m;
char s

;
double g

;

int num (char ss[]); //判断与前边的字符串是否相等
void Floyd ();

int main ()
{
    char s1
, s2
;
    double t;
    int flag = 1;

    while (cin >> n, n)
    {
        getchar ();
        for (int i=1; i<=n; i++)
            gets (s[i]);

        cin >> m;
        memset (g, 0, sizeof (g));

        for (int i=1; i<=m; i++)
        {
            cin >> s1 >> t >> s2;
            g[num(s1)][num(s2)] = t;
        }

        cout << "Case " << flag++ << ": ";
        Floyd ();
    }
    return 0;
}

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

                if (i == j && g[i][j] > 1)//如果交换后金额大于之前
                {
                    puts ("Yes");
                    return;
                }
            }
        }
    }
    puts ("No");
}

int num (char ss[])
{
    for (int i=1; i<=n; i++)
        if (!strcmp (ss, s[i]))
            return i;
}