Colleagues, how about the following?
At each start point, the total number of leaving elephants increases by
one; at each end point, it decreases by one. We:
1. form a vector of pairs: {5, +1}, {10, -1}, {6, +1}, {15, -1}, {2, +1},
{7, -1} ... -- this takes O(N) time and O(N) additional spac
Monish, please take a look at a similar problem about poor elephants in
the neighboring topic started today. I believe the problem has a similar
solution. Each start point increases the no. of active intervals by one;
each end point decreases it. So, we do the following:
1. Convert the inpu
How about the following:
1. Think every cell on a board occupied by an "O" is a vertex in a
non-directed graph. Add a special vertex called "OuterWorld".
2. Connect all neighboring vertices with edges, based on how you define
"surrounded by X-es". E.g. in your picture, diagonals are not allowed
Start at this one: http://en.wikipedia.org/wiki/Alpha%E2%80%93beta_pruning
-IB
On Friday, August 2, 2013 1:56:14 AM UTC-4, Naren s wrote:
>
>
> Can anyone help me solve this problem.
>
> http://poj.org/problem?id=1085
>
>
> --
> Regards,
> *Narayanan S*
>
>
--
You received this message becaus
Hey, isn't it as simple as:
1. Start at sum of the first K elements.
2. Move the "window" of size K by one element to the right at each step.
Subtract the left, add the right => get the new sum. Keep the maximum.
This executes in O(n) steps, requires O(1) memory. The code in C++ seems to
be tr
