Arbitrage

                                  Arbitrage

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

题目大意：给定N种货币及它们之间的汇率，问能否经过一系列交换之后让货币升值？

思路:最短路的变形，只需要把权值+变为 * 就可以，字符串处理要注意。

code：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;char a[35][50];double b[35][35];int n;int floyd(){    for(int k=1;k<=n;k++)      for(int i=1;i<=n;i++)        for(int j=1;j<=n;j++)          if(b[i][j] < b[i][k]*b[k][j])            b[i][j] = b[i][k] * b[k][j];    for(int i=1;i<=n;i++)        if(b[i][i] > 1.0) return 1;    return 0;}int search(char ch[]){    for(int i=1;i<=n;i++)      if(strcmp(ch,a[i]) == 0) return i;}int main(){    int tot=0;    while(~scanf("%d",&n) && n)    {        memset(b,0,sizeof(b));        for(int i=1;i<=n;i++)            scanf("%s",a[i]);        int m;        scanf("%d",&m);        char c1[50],c2[50];        double rate;        for(int i=1;i<=m;i++)        {            scanf("%s %lf %s",c1,&rate,c2);            b[search(c1)][search(c2)] = rate;        }        printf("Case %d: ",++tot);        if(floyd())            printf("Yes\n");        else            printf("No\n");    }    return 0;}