62 lines
2.5 KiB
Markdown
62 lines
2.5 KiB
Markdown
---
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id: SRQ
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title: "Static Range Queries"
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author: Benjamin Qi
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description: Range queries for any associative operation over a static array.
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---
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Given a static array $A[1],A[2],\ldots,A[N]$, you want to answer queries in the form $A[l]\ominus A[l+1]\ominus \cdots \ominus A[r]$ where $\ominus$ denotes any associative operation.
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This can be done in $O(1)$ time each with $O(N\log N)$ time preprocessing.
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## [Range Minimum Query](https://en.wikipedia.org/wiki/Range_minimum_query)
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First we'll consider the special case when $\ominus$ denotes `min`.
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- [CSES Range Minimum Queries I](https://cses.fi/problemset/task/1647)
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### Tutorial
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- CPH 9.1
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- [PAPS 11.2.2](https://www.csc.kth.se/~jsannemo/slask/main.pdf)
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- [cp-algo RMQ](https://cp-algorithms.com/sequences/rmq.html)
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- [cp-algo Sparse Table](https://cp-algorithms.com/data_structures/sparse-table.html)
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<optional-content title="Preprocessing in O(N) Time">
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- [CF: $O(1)$ Query RMQ with $O(N)$ build](https://codeforces.com/blog/entry/78931)
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</optional-content>
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## Divide & Conquer
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**Divide & conquer** can refer to many different techniques. In this case, we use it to answer $Q$ queries offline in $O((N+Q)\log N)$ time.
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Suppose that all queries satisfiy $L\le l\le r\le R$ (initially, $L=1$ and $R=N$). Letting $M=\left\lfloor \frac{L+R}{2}\right\rfloor$, we can compute
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$$
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lef[l]=A[l]\ominus A[l+1]\ominus \cdots \ominus A[M]
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$$
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for all $L\le l\le M$ and
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$$
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rig[r]=A[M+1]\ominus A[M+2] \ominus \cdots\ominus A[r]
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$$
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for each $M< r\le R$. Then the answer for all queries satisfying $l\le M< r$ is simply $lef[l]\ominus rig[r]$ due to the associativity condition. After that, we recurse on all query intervals completely contained within $[L,M]$ and $[M+1,R]$ independently.
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Actually, this can be adjusted to answer queries online in $O(1)$ time each. See my implementation [here](https://github.com/bqi343/USACO/blob/master/Implementations/content/data-structures/Static%20Range%20Queries%20(9.1)/RangeQuery.h).
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<optional-content title="Faster Preprocessing">
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A data structure known as **sqrt-tree** can speed up preprocessing time to $O(N\log \log N)$.
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- [CF Blog Pt 1](http://codeforces.com/blog/entry/57046)
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- [CF Blog Pt 2](http://codeforces.com/blog/entry/59092)
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</optional-content>
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## Problems
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- [Codechef - Product on Segment](https://www.codechef.com/problems/SEGPROD)
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- [DMOJ - Continued Fractions](https://dmoj.ca/problem/dmopc19c7p4)
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- [USACO Plat - Non-Decreasing Subsequences](http://www.usaco.org/index.php?page=viewproblem2&cpid=997)
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- [JOI - Secret](https://oj.uz/problem/view/JOI14_secret)
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