My English is poor as a Chinese...

A few days ago I post an idea at a Chinese Online Judge named Luogu.This is my first idea created by myself to solve RMQ problem. Unfortunately it's Chinese.The basic idea is divide the array into $$$n^\frac{4}{7}$$$ part,each part has a length of $$$n^\frac{3}{7}$$$.We name it as a "big block".Then we divide each block into $$$n^\frac{2}{7}$$$ part,each part has a length of $$$n^\frac{1}{7}$$$.Thus it can has a time complexity of $$$O(n^\frac{8}{7})$$$.It is quicker than ST Table both in theory and practice.

And then I create second idea.As we all know there is an algo called Four Russian.It can solve RMQ problem at a time complexity $$$O(n\log\log n)$$$.We found when we finish initialization,we are actually solve an RMQ problem but a smaller scale.Then we use Four Russian to this smaller RMQ problem,and then our time complexity become $$$O(n\log\log\log n)$$$.And we do it again while $$$n\neq 1$$$,it can do it in a time complexity of $$$O(n\log^*n)$$$.This can be called Sqrt Tree Set Four Russia(I like to call it as Log Tree)

This is a very outstanding algorithm but it's not enough.To improve it,we can set up many data structure and the .We set $$$T(m,n)$$$ to be the complexity of the $$$m$$$-layer data structure.We build a Sqrt Tree with $$$B=T(m-1,n)$$$ and let it be the $$$m$$$-layer data structure.Between block and block we use the $$$m-1$$$-layer data structure to solve the problem and in a block we use next layer of this data structure,then,as Base Four Russian Multiple Sqrt Tree(BFRMST) I called,finish.It is an algo of $$$O(n\alpha(n))$$$.

Ah somebody tell me that I need to talk about Four Russian,so I'll tell it.

The algo Four Russian can solve RMQ problem in a time complexity of $$$O(n\log \log n)-O(1)$$$.

It is work like that:we divide the array to $$$O(\frac{n}{\log n})$$$ parts and each parts got $$$O(\log n)$$$ items.

We set up an ST Table to uphold the maximum value between block and block,and between each block.

Obviously we can get answer using these.

The time complexity is $$$O(\frac{n}{\log n}\log n\log \log n)=O(n\log \log n)$$$.

FeS2_qwq independently invented this algorithm.

Lumos_QwQ and zhaoyuebo tested the theoretical correctness of this algorithm.

zyb_txdy coding for it.

You can find the algo Sqrt Tree on here.

Fun fact:when I finish it,CF crashed I use mirror to post it.

Auto comment: topic has been updated by FeS2_qwq (previous revision, new revision, compare).Auto comment: topic has been updated by FeS2_qwq (previous revision, new revision, compare).Auto comment: topic has been updated by FeS2_qwq (previous revision, new revision, compare).Auto comment: topic has been updated by FeS2_qwq (previous revision, new revision, compare).Auto comment: topic has been updated by FeS2_qwq (previous revision, new revision, compare).How can I delete these strange auto comment

when u're updating, uncheck the auto comment box

too classic

Really?All right I think it's my own invention...

Auto comment: topic has been updated by FeS2_qwq (previous revision, new revision, compare).Auto comment: topic has been updated by FeS2_qwq (previous revision, new revision, compare).Auto comment: topic has been updated by FeS2_qwq (previous revision, new revision, compare).Auto comment: topic has been updated by zhaoyuebo (previous revision, new revision, compare).Auto comment: topic has been updated by zhaoyuebo (previous revision, new revision, compare).We found when we finish initialization,we are actually solve an RMQ problem but a smaller scale.I cannot understand what this sentence is about. AFAIK The Method of Four Russians only support delta-1 RMQ.

Maybe you should contain an implementation in this post?

I'll upd when I have time.

What you said is a improvement of Four Russian.

Thanks for advice,upvote u.

could you pls tell me where you posted this algorithm on Luogu?

It is not this,is the $$$O(n^\frac{8}{7})$$$ one.

this.

You construct the data structure in $$$O(n^{\frac 8 7})$$$ but how would you answer a query better than $$$O(n^{\frac 2 7})$$$ this way?

Ah,let me think.

That is not important(just kidding).

All right I think I need to spend some time thinking it seriously.

Aha my code is O(1) to answer a query. You can initialize the answer in a block and between every block.

Code

Oh ok, i see. Nice idea !

You mentioned the sqrt tree blog. There is actually a second part: https://mirror.codeforces.com/blog/entry/59092

About update: sometime I will think of it carefully and write it here. It is seem to be $$$O(\alpha(n) \log n)$$$.Can you please elaborate a little more about how to achieve such update complexity?

Maybe update it in a sqrt-tree way?

But I didn't think carefully of it is $$$O(\alpha(n)\log n)$$$ or $$$O(\frac{n\alpha(n)}{\log n})$$$.

The second part of the blog mentions that sqrt-tree updates only in $$$O(\sqrt{n})$$$ (and provides a way to do this). So this is an interesting question.

Ah I know.It is $$$O(\alpha(n)\log n)$$$.

I'll update Four Russian in 3 days and update in a week.

Maybe I'll resolve it to two or three blogs.

Correct:It can be solve in $$$O(\log n)$$$.

Oh I get it wrong.It cannot update (-:

Mainly because we cannot update an ST Table

I finish Four Russian.

seems the same as the algorithm described in this paper https://mirror.codeforces.com/blog/entry/131221?#comment-1167384