89 lines
3.2 KiB
Text
89 lines
3.2 KiB
Text
---
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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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frequency: 1
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---
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import { Problem } from "../models";
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export const metadata = {
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problems: {
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rmqSample: [
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new Problem("YS", "Static RMQ", "staticrmq", "Easy", false, [], "equivalent to [CSES Range Minimum Queries I](https://cses.fi/problemset/task/1647)"),
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],
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diviSample: [
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new Problem("ojuz", "JOI Secret", "JOI14_secret", "Easy", false, [], ""),
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],
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general: [
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new Problem("CC", "Product on Segment", "SEGPROD", "Normal", false, [], ""),
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new Problem("DMOJ", "Continued Fractions", "dmopc19c7p4", "Hard", false, [], ""),
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new Problem("Plat", "Non-Decreasing Subsequences", "997", "Very Hard", false, [], ""),
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],
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}
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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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With $O(N\log N)$ time preprocessing, we can get $O(1)$ queries.
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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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<Problems problems={metadata.problems.rmqSample} />
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### Resources
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<Resources>
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<Resource source="CPH" title="9.1 - Minimum Queries" starred>diagrams</Resource>
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<Resource source="PAPS" title="11.2.2 - Sparse Tables" starred>code</Resource>
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<Resource source="cp-algo" title="RMQ" url="sequences/rmq.html"></Resource>
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<Resource source="cp-algo" title="Sparse Table" url="data_structures/sparse-table.html"></Resource>
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</Resources>
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<Optional title="Faster Preprocessing">
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[CF: $O(1)$ Query RMQ with $O(N)$ build](https://codeforces.com/blog/entry/78931)
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</Optional>
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### Implementation
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<IncompleteSection />
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## Divide & Conquer
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<Problems problems={metadata.problems.diviSample} />
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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 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>
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### Implementation
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<IncompleteSection />
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## Problems
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<Problems problems={metadata.problems.general} /> |