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Need idea to solve LightOj -1089
By darkprinx, 10 years ago, In English

i started to thinking about segment tree to solve this problem.

LightOj -1089

But after i noticed the range [0, 10^8], i just stucked there and couldn't managed how to solve it. please help ... thnx in advance :) :)

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10 years ago, # |
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But after i noticed the range [0, 10^8], i just stucked there and couldn't managed how to solve it

You have to think about the amount of distinct numbers you have (at most: 2 × 50 000 for the ranges, plus 50 000 for the query points).

For example, if I give you these ranges: [1, 60 000 000], [40 000 000, 90 000 000] and these query points: 1 000 000, 50 000 000, 100 000 000, then you can reduce the problem (without changing the final answer) to a smaller problem. In this case, the ranges become: [1, 5], [3, 6], and the query points become: 2, 4, 7. For "both problems", the answers are 1, 2, 0.

P.S. note that (once you solved the "reduction" part) you can solve this problem also by using a binary indexed tree (with a trick I explained here) which can be easier than a range tree.

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    7 years ago, # ^ |
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    I don't understand how to do the reduction part. How you reduce this one -> [1, 60 000 000], [40 000 000, 90 000 000] and these query points: 1 000 000, 50 000 000, 100 000 000 to this one [1, 5], [3, 6], and the query points become: 2, 4, 7 Both are quite different. isn't it?

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      7 years ago, # ^ |
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      To compress numbers you can replace a certain number with its index in sorted array, for example: [1, 6, 28, 5], sorted array is [1, 5, 6, 28], so you will get new array [0, 2, 3, 1], don't forget that equal numbers must be equal in new array too.

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        7 years ago, # ^ |
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        Thanks for Your Reply. I understand now.

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6 years ago, # |
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The problem can easily be solved using upper_bound and lower_bound . Just sort up the left points in a vector and the right point in another vector

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