10797 - Peaceful Sharing

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Larry
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10797 - Peaceful Sharing

Post by Larry »

Is there a simpler way to solve this without calculating the median of the arrangements of the duals? It's O(n^2), but is that the solution the problemsetter was looking for (the implementation is long..)? Thanks for any hints.

shahriar_manzoor
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hmm

Post by shahriar_manzoor »

I don't know what length will u consider long :)

My faster solution is 140 Lines
My slower (three times) is 100 lines
Kisman's solution is 40 lines (With STL) which is also quite slow but runs within time limit.

Cho
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Post by Cho »

my solution is less than 70 lines, without using any library except stdio.h.

shahriar_manzoor
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Re: 10797 - Peaceful Sharing

Post by shahriar_manzoor »

Larry wrote:Is there a simpler way to solve this without calculating the median of the arrangements of the duals? It's O(n^2), but is that the solution the problemsetter was looking for (the implementation is long..)? Thanks for any hints.
I also forgot to reply the actual question. Yes I used median of arrangement as my idea to solve this problem. Kisman's idea was similar I think.

Christian Schuster
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Post by Christian Schuster »

My solution has about 100 lines and uses a tiny bit of STL. I use an iterative approach, improving the selected mines in an alternating manner... And it seems to be quite fast. :D

Destination Goa
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Post by Destination Goa »

I've got an idea of N*log^2(N) solution. If it works, I'll write it here.

P.S: Was that authors' intention of N^2 solution or they just didn't know how to code a better way for the problem they've put, but are aware of its existence?
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Destination Goa
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Post by Destination Goa »

Well, my method didn't work (tried binary search over convex hull nodes). Though, there still might be another one which consider inner points as well. I was unable yet to find any n*log(n) ordering.

By the way, if the intentional complexity was O(N^2), so you wanted to prevent people from sending O(N^3), why did you set limit to 10'000, but not to 2000? 10'000 is not better than 2000 as O(N^3) blocker, but it is very confusing limit as O(N^2) allower.
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Destination Goa
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Post by Destination Goa »

Also I won't be surprised if O(N^2) testing of the 1st set via 2nd set will differ from the opposite approach (testing 2nd set via 1st set) 5 or 6 times on time. E.g. TL will turn into AC once you make x=-x. Fair square limit of 2000-3000 won't allow such weird thing to happen.
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shahriar_manzoor
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hmm

Post by shahriar_manzoor »

I don't think the judge solution is O(N^2).

I have two solutions one was O(N^2) and modifying it slightly I made a better solution. But I allowed both the better and the O(N^2) to pass. I did not want a very bad O(N^2) to pass that is why N=10000. And I don't understand what is your problem if N=10000 and not 2000.

Destination Goa
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Post by Destination Goa »

Shahriar,
I don't think the judge solution is O(N^2).
If you are the author, they why don't you know? :) Or you're not?

The problem is that whenever I see N=10'000 or greater, I seek for N*log^k(N) solution. I think that problems are intended to demand some asymptotics, but not optimizing N^2 (or N^3) to make it AC. Because for any such "solution" there is always a test which makes its worst. And all those "optimizatiosn" just increase the contanst, and you get a fair TL. Limits are a way to set an upper limit for demanded asymptotic.

Perhaps I am just too old problem solver, and what before was 1'000 now is 10'000. Well, never mind.

THE QUESTION:
Is there a way to build nested convex hulls of set of points better than at O(N^2)? I.e. once the outer hull is built, delete all its nodes and do it again until no points left (innermost hull may contain 1 or 2 points). If such algoritmh exists, there is possibility of logarithm for this problem.
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Destination Goa
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Post by Destination Goa »

Yes, here it is. Called "Onion-peeling". http://cgm.cs.mcgill.ca/~orm/ontri.html
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shahriar_manzoor
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hmm

Post by shahriar_manzoor »

To be specific the judge solution is O(cN), where c is a constant of value around 100. "I don't think..." was another way to say "is not".

I never thought of convex hull while setting this problem so cannot comment on your issue.

Destination Goa
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Post by Destination Goa »

No, onion-peeling won't work either. Well, it will give correct answer, but we'll have logarithm only for layer length, not on number of layers. So, if each layer has 3 nodes, we'll stay at O(n1*n2) = O(N^2).

What do you mean by C=100? CPU cycles, some fixed time or what? Or is this some sort of function which is 100 for N<=10'000? Would it remain 100 for N<=1'000'000?
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shahriar_manzoor
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hmm

Post by shahriar_manzoor »

I used bisection in this problem. C is the number of iteration required in bisection to find the answer in the worst case. When N=10000000 C will not change.

Destination Goa
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Post by Destination Goa »

Uh. I read one of your 1st posts incorrectly. You said that you also seek for the median, but you didn't say that you also perform it at O(N^2) :)

Anyway, my question about C remains.
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