120 lines
4.1 KiB
Text
120 lines
4.1 KiB
Text
---
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id: merging
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title: "Small-To-Large Merging"
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author: Michael Cao
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prerequisites:
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- Silver - Depth First Search
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- Gold - Point Update Range Sum
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description: "?"
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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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sam: [
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new Problem("CSES", "Distinct Colors", "1139", "Intro", false, ["Merging"]),
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],
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general: [
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new Problem("CF", "Lomsat gelral", "contest/600/problem/E", "Normal", false, ["Merging"]),
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new Problem("Plat", "Promotion Counting", "696", "Normal", false, ["Merging", "Indexed Set"], ""),
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new Problem("Gold", "Favorite Colors", "1042", "Hard", false, ["DSU"], "Small to large merging is mentioned in the editorial, but we were unable to break solutions that just merged naively."),
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new Problem("Plat", "Disruption", "842", "Hard", false, ["Merging"]),
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]
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}
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};
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## Additional Reading
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<resources>
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<resource source="CPH" title="18.4 - Merging Data Structures"></resource>
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<resource source="CF" title="Arpa - Sack (DSU on Tree)" url="blog/entry/44351"></resource>
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<resource source="CF" title="tuwuna - Explaining DSU on Trees" url="blog/entry/67696"></resource>
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</resources>
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## Merging Sets
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<problems-list problems={metadata.problems.sam} />
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Let's consider a tree rooted at node $1$, where each node has a color.
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For each node, let's store a set containing only that node, and we want to merge the sets in the nodes subtree together such that each node has a set consisting of all colors in the nodes subtree. Doing this allows us to solve a variety of problems, such as query the number of distinct colors in each subtree. Doing this naively, however, yields a runtime complexity of $O(N^2)$.
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However, with just a few lines of code, we can significantly speed this up. Note that [swap](http://www.cplusplus.com/reference/utility/swap/) exchanges two sets in $O(1)$ time.
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```cpp
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if(a.size() < b.size()){ //for two sets a and b
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swap(a,b);
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}
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```
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By merging the smaller set into the larger one, the runtime complexity becomes $O(N\log N).$ ?? (Ben - this is false)
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### Proof
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When merging two sets, you move from the smaller set to the larger set. If the size of the smaller set is $X$, then the size of the resulting set is at least $2X$. Thus, an element that has been moved $Y$ times will be in a set of size $2^Y$, and since the maximum size of a set is $N$ (the root), each element will be moved at most $O(\log N$) times leading to a total complexity of $O(N\log N)$.
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<spoiler title="Full Code">
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```cpp
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#include <bits/stdc++.h>
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using namespace std;
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const int MX = 200005;
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vector<int> adj[MX]; set<int> col[MX]; long long ans[MX];
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void dfs(int v, int p){
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for(int e : adj[v]){
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if(e != p){
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dfs(e, v);
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if(col[v].size() < col[e].size()){
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swap(col[v], col[e]);
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}
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for(int a : col[e]){
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col[v].insert(a);
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}
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col[e].clear();
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}
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}
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ans[v] = col[v].size();
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}
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int main() {
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ios::sync_with_stdio(false);
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cin.tie(0);
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int n; cin >> n;
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for(int i = 0; i < n; i++){
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int x; cin >> x;
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col[i].insert(x);
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}
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for(int i = 0; i < n - 1; i++){
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int u,v; cin >> u >> v;
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u--; v--;
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adj[u].push_back(v); adj[v].push_back(u);
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}
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dfs(0,-1);
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for(int i = 0; i < n; i++){
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cout << ans[i] << " ";
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}
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}
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```
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</spoiler>
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## Generalizing
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A set doesn't have to be an `std::set`. Many data structures can be merged, such as `std::map` or `std:unordered_map`. However, `std::swap` doesn't necessarily work in $O(1)$ time; for example, swapping two [arrays](http://www.cplusplus.com/reference/array/array/swap/) takes time linear in the sum of the sizes of the arrays, and the same goes for indexed sets. For two indexed sets `a` and `b` we can use `a.swap(b)` in place of `swap(a,b)` (documentation?).
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<info-block title="Challenge">
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Prove that if you instead merge sets that have size equal to the depths of the subtrees, then small to large merging does $O(N)$ insert calls.
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(be specific about what this means?)
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</info-block>
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## Problems
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<problems-list problems={metadata.problems.general} />
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