Arbitrage
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 21381 | Accepted: 9110 |
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.
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.
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
Sample Output
Case 1: Yes Case 2: No
Source
题解：
题意是给你各个货币之间的的转换率(单向），然后问你能否经过一些转换使得货币变多，就是1美元可以换6元人民币，6元人民币可以换100日元，每80日元可以换1美元，然后你的1美元就可以通过这个变换变成1.25美元，那么就输出Yes，如不能升值，则输出No。
可以看出来，这道题询问的是能否再回到这个点是>1，则我们就可以用floyd来求，因为这道题问的是多源路，所以spfa和dijkstra会很慢。
注意，这道题需要用double
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
using namespace std;
map <string, int> a;
double d[60][60], ans = 1, n;
void floyd()
{
for(int k = 1;k <= n;k ++)
{
for(int i = 1;i <= n;i ++)
{
for(int j = 1;j <= n;j ++)
{
if(d[i][j] < d[i][k] * d[k][j])
d[i][j] = d[i][k] * d[k][j];
}
}
}
}
int main()
{
while(cin>>n)
{
if(n == 0)
break;
memset(d, 0, sizeof(d));
a.clear();
for(int i = 1;i <= n;i ++)
{
string aa;
cin>>aa;
a.insert(make_pair(aa, i));
d[i][i] = 1;
}
int m;
cin>>m;
for(int i = 1;i <= m;i ++)
{
string aa, b;
double s;
cin>>aa>>s>>b;
d[a[aa]][a[b]] = s;
}
floyd();
bool flag = 0;
for(int i = 1;i <= n;i ++)
{
if(d[i][i] > 1)
{
flag = 1;
break;
}
}
if(flag)
cout<<"Case "<<ans++<<": Yes"<<endl;
else
cout<<"Case "<<ans++<<": No"<<endl;
}
return 0;
}