Hey, I was trying to solve this problem : http://www.usaco.org/index.php?page=viewproblem2&cpid=622

I came up with following approach which is kind of messy:

• States:

1. Starting position (the first door that is open) (i)
2. Ending position (the current last door that will be opened) (j)
3. Count of doors already open before (k)

• Transition for all x < j: dp[i][j][k] = min(dp[i][x][k-1] + {something here}, dp[i][j][k-1])

{something here} = sum 1. for all y such that j-x+y >=0, x+(j-x+y)/2 , sum+= A[x+(j-x+y)/2]*(j-x+y+1) 2. for all y such that y < (j+(i+1))/2 , sum+=A[y]*(y-j); 3. for all y such that y > (j+(i+1))/2 and y < n , sum+=A[y]*(i+(n-y)); 4. for all y such that y>=0 && y<i , sum+=A[y]*(y+1);

similar numbers will be subtracted from the solution.

What I am doing is, when we add new door/room, I take half the cows between last door and current door and add them to the solution.

Here is the actual solution of the problem : http://www.usaco.org/current/data/sol_cbarn2_gold_feb16.html

Can anyone explain me the following:

1. If my approach correct ? (Forgetting the time constraints for now)
2. How can this solution be obtained by optimising my solution ?
3. How you came up to the solution ?