I think a sequence of vertices in (counter)clockwise order with
constant time access by index will well suffice. Below I assume the
counterclockwise order, which I prefer in most cases.
My approach is to first split the polygon into two chains of vertices
at two extremes with respect to the
you can find it here the World Funniest Video at
http://www.supperlaffn.blogspot.com
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Hi,
could anybody help me to solve this problem
http://acm.zju.edu.cn/show_problem.php?pid=1002
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This is easy BackTrack problem ,I think
On 2/20/07, ramtin [EMAIL PROTECTED] wrote:
Hi,
could anybody help me to solve this problem
http://acm.zju.edu.cn/show_problem.php?pid=1002
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Won't brute force does the trick? Looks like it's at most 4x4 (correct
me if I'm wrong). You could play around with bitmasks probably, ... treat
each cell as one bit, ... probably also worthwhile to see what the bitmask
corresponds to the controlled cells if a rook is placed in certain
I'm sorry but I couldn't understand your code
and I think we can't use BackTrack because it is useful for problem
that we want to find a good leaf
but in this problem we must check all possible combinations
On Feb 20, 11:34 pm, Lego Haryanto [EMAIL PROTECTED] wrote:
Won't brute force does
Ok, ... it's easier to view it this way ... consider a maximum 4x4 board.
So, this is the same as 16 cells. In each cell, obviously it could either
be 0 or 1, ... 0 being empty, and 1 being occupied.
Just looking at the fact above, ... you can see we can have at most 2^16
combinations or
