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：
因为题中货币汇率的转换使用的是乘法，而与小于1的数相乘是会导致原来的数变小，所以此题相当于是有负权边的题目，那么Dijkstra算法是不能使用的。题目问套利是否存在，因此每种货币是否存在套利都应该考虑，Floyd算法的耗时不会太长。然而题目中给出的任意两种货币不一定能够互相转换，那么不能够转换的话我们用0表示，使用Floyd算法求每两种货币能够互相转换的最大汇率，那么最后我们判断每种货币转换为自己时的汇率是否大于1，是的话说明存在套利，否的话则说明本货币不存在套利。注意本题是单向图。。。
程序代码：
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#define max(a,b)(a>b?a:b)
#define N 50
char str[N][N];//存储货币名称
double map[N][N];//存储汇率
int n,m;
int count(char s[])//计算字符串编号
{
int i;
for(i=0;i<n;i++)
{
if(strcmp(s,str[i])==0)//匹配到相符的字符串
return i;
}
return -1;
}
int Floyd()
{
int i,j,k;
for(k=0;k<n;k++)
{
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
map[i][j]=max(map[i][j],map[i][k]*map[k][j]);
}
}
}
for(i=0;i<n;i++)//判断是否存在套利
{
if(map[i][i]>1)//货币转换成自身大于1则存在套利
return 1;
}
return 0;
}
int main()
{
int i,ans=1;
char s1[N],s2[N];
double x;
while(~scanf("%d",&n))
{
if(n==0)
break;
memset(map,0,sizeof(map));
for(i=0;i<n;i++)
scanf("%s",str[i]);
scanf("%d",&m);
for(i=0;i<m;i++)
{
scanf("%s%lf%s",s1,&x,s2);
map[count(s1)][count(s2)]=x;
}
printf("Case %d: ",ans++);
if(Floyd()==1)
printf("Yes\n");
else
printf("No\n");
}
return 0;
}
