Arbitrage POJ - 2240（金钱转换）

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：还是金钱转换问题，转换方式：将一个货币单位转换成同一货币的一个以上单位。例如，假设1美元买0.5英镑，1英镑买10法郎，1法郎买0.21美元。然后，通过转换货币，一个聪明的交易者可以从1美元开始，买入0.5×10×0.21＝1.05美元，获利5%。求最后是否能盈利，这道题没有手续费问题，但是对于输入的字符串（两个地市之间的金钱转换额）转换成矩阵，因为没有初始金钱多少，所以都设成1.

#include
#include
#include
double r[1000][1000];//汇率
char s[1000][1000];
int n,m;
int zhuan(char buf[])
{
    int i;
    for(i=0; iif(strcmp(buf,s[i])==0)
            return i;
return 0;
}
void Input()
{
    char buf1[1000],buf2[1000];
    double rate;
    for(int i=0; iscanf("%s",s[i]);
    }
    memset(r,0,sizeof(r));
    for(int i=0; i1;//初始钱的金额设成1
    }
    scanf("%d",&m);
    for(int i=0; iscanf("%s %lf %s",buf1,&rate,buf2);
        r[zhuan(buf1)][zhuan(buf2)] = rate;//将两城市之间的汇率关系转换成矩阵
    }
}
int Floyd()
{
    int i,j,k;
    for(  k = 0; k < n; k ++)
        for(  i = 0; i < n; i ++)
            for( j = 0; j < n; j ++)
                if( r[i][j] < r[i][k] * r[k][j])
                    r[i][j] = r[i][k] * r[k][j];
    for( i = 0; i < n; i ++)
        if( r[i][i] > 1)
            return 1;
    return 0;
}
int main()
{
    int cas = 0;
    while( scanf( "%d", &n), n)
    {
        Input();
        printf( "Case %d: ", ++ cas);
        int ok = Floyd();
        if(ok) 
        printf( "Yes\n");
        else 
        printf( "No\n");
    }
    return 0;
}