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frostcloud's blog

Help me to make a data structure
By frostcloud, history, 6 days ago, In English

Create a data structure, that can:

  1. Add/subtract a number on a segment;
  2. Count amount of negative numbers on a segment.

How to solve it for $$$O(\sqrt{n})$$$ / $$$O(\log^2{n})$$$ / $$$O(\log{n})$$$ per query?

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6 days ago, # |
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I think what you are looking for is lazy segment tree + merge sort tree. time complexity is log^2 n per query

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    6 days ago, # ^ |
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    I know about lazy segment tree, but don't know about merge sort tree.

    What is the Merge Sort Tree? How is it implemented? And the main question — how can i update it on some range?

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      6 days ago, # ^ |
      ← Rev. 3   Vote: I like it -26 Vote: I do not like it

      From Chat GPT: The merge sort tree is a type of segment tree that merge sorts a randomly-generated array of length $$$2 \cdot 10^5$$$ and element boundary $$$1 \le a_i \le 10^9$$$ after each query. Because of this property, it is very inefficient, and should never be used for non-educational purposes.

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      6 days ago, # ^ |
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      https://www.geeksforgeeks.org/merge-sort-tree-for-range-order-statistics/ learn this first.

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        6 days ago, # ^ |
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        I found out that I am wrong you can't make range updates using merge sort tree and then count amount of negative numbers on a segment but you can do point updates.

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          6 days ago, # ^ |
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          that's why it is said , think before you act.

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6 days ago, # |
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sqrt decomposition my beloved ($$$O(\sqrt[]{n}log n)$$$ sol so might be a bit slow)

divide the array into $$$\sqrt[]{n}$$$ segments of length $$$\sqrt[]{n}$$$ (and store them twice, one of them is the sorted version)

add/subtract query would take $$$O(\sqrt[]{n} log n)$$$:

for each block store a value which is how much the entire segment is affected by (we'll call it $$$a_i$$$), if the query affects the entire block then just update $$$a_i$$$

if it only affects it partially, then subtract/add those elements and build the sorted array again

for searching negative numbers -> for partial segments just loop through them, for full segments: binary search to find the number of values smaller than $$$-a_i$$$ (also $$$O(\sqrt[]{n}log n)$$$)

obviously it's not the runtime you're looking for but uhm...

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    6 days ago, # ^ |
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    I have also thought about it. This is a good idea, but, unfortunately, the complexity is too high.

    Maybe I can do some heavy optimizations, so that it will work for $$$n = queries = 10^5$$$?

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      5 days ago, # ^ |
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      what is the source of this problem btw? did you come up with it by yourself, or did you reduce some problem on an online judge to this?

      and yes, $$$n\sqrt{n}\text{ log n} \approx 5 \cdot 10^8$$$ operations are probably possible if the code is optimized enough (assuming the time limit isn't too tight)

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        5 days ago, # ^ |
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        yes, good question. because once I reduced a problem to this problem, and thought n-sqrt-log was too slow. But then after another observation, you can reduce it to the above problem but all values are non-negative; and count zeros instead of negatives

        then you can do lazy seg with min and number of mins

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    5 days ago, # ^ |
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    If you use segments of length sqrt(NlogN) and split+merge when doing updates to a part of a segment, then the operations work in sqrt(NlogN).

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      4 days ago, # ^ |
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      Wow, this makes a lot of sense. How does one go about optimizing block sizes like this? Do you have to just "see" it, or are there some ideas that help?

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        4 days ago, # ^ |
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        Write down the cost of operations.

        If you're able to update/query one block of size N in f(N) and update/query a block partially in g(N) then the cost is (N / X) * f(X) + 2 * g(X). Optimizing that expression we get that X must be O(sqrt(NlogN)) if f(X) = logN and g(X) = X.

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5 days ago, # |
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You can use the lazy segment tree + treap (or any other balanced tree) to solve it in $$$\mathcal O(\log^2 n)$$$ per query. In implementation, you can build a balanced tree for each node in segment tree, and it controls the whole subtree of that node (and as you know, a subtree in segment tree means a interval). Evidently, every node in segment tree will be in $$$\mathcal O(\log n)$$$ balanced trees, so it has a right complexity.

To make the change, you can do it like the simple lazy segment tree. So you can maintain a value $$$add$$$ in segment tree which means "how much this subtree of the node (a interval in fact) add/subtract". When you query, you also just do it like the simple lazy segment tree, but on each node, you should query the rank of $$$-add$$$, because now $$$-add$$$ is $$$0$$$ in fact.

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    5 days ago, # ^ |
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    pretty sure this fails for similar reasons why merge sort tree fails: when returning back up, how do you merge 2 updated treaps in O(logn)?

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      5 days ago, # ^ |
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      emm, you're right. the tag is unable to turn around the hopeless situation of this solution.

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5 days ago, # |
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There exist a variant of this problem which only deals with subtractions: https://oj.vnoi.info/problem/bedao_m23_e (obviously $$$O(n\sqrt{n}log(n))$$$ does not work here)

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    5 days ago, # ^ |
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    By the way, $$$O(n\sqrt{n}log(n))$$$ does work for subtask 2, which matches your constraints.

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