PROBLEMS (the main part of each contest) are IOI stylish

Yes, I also feel really sad about that.

my solution:

the state is dp[mask] where mask value is at most (2^n)-1 (I started counting form 0)

in this mask if the iTH bit was ON this means the iTH glass is not empty.

if the number of the turned on bits is k the answer is zero.

or you will apply two "for" loops on this mask..

let's say:

for(int i=0;(1<<i)<=mask;i++)

for(int j=0;(1<<j)<=mask;j++)

Now if the iTH bit or the jTH bit is off or i equals to j ,don't do any thing.

or you will try to pure the water from iTH glass to the jTH glass and the answer is a[i][j]+dp[mask-(1<<i)]

where a[i][j] is the cost to pure the water from the iTH glass to the jTH glass and you will turn off the iTH bit because the iTH glass in empty after that

my code

I put factor "on" in the solve function because I don't want to count the turned on bits everytime (this doesn't effects on the memorization because every mask has unique number of turned on bits).

I hope this information is useful :D

sorry for my bad English.

My really easy solution:

http://pastebin.com/4v3sZvk7