usaco-guide/content/6_Plat/Merging.mdx

---
id: merging
title: "Small-To-Large Merging"
author: Michael Cao, Benjamin Qi
prerequisites: 
 - Silver - Depth First Search
 - Gold - Point Update Range Sum
description: "?"
frequency: 1
---

import { Problem } from "../models";

export const metadata = {
  problems: {
    sam: [
      new Problem("CSES", "Distinct Colors", "1139", "Intro", false, ["Merging"]),
    ],   
    general: [
      new Problem("CF", "Lomsat gelral", "contest/600/problem/E", "Normal", false, ["Merging"]),
      new Problem("Plat", "Promotion Counting", "696", "Normal", false, ["Merging", "Indexed Set"], ""),
      new Problem("Plat", "Disruption", "842", "Normal", false, ["Merging"]),
      new Problem("POI", "Tree Rotations", "https://szkopul.edu.pl/problemset/problem/sUe3qzxBtasek-RAWmZaxY_p/site/?key=statement", "Normal", false, ["Merging", "Indexed Set"], ""),
      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. Alternatively, just merge linked lists in $O(1)$ time."),
    ],
    treeRot: [
      new Problem("POI", "Tree Rotations 2", "https://szkopul.edu.pl/problemset/problem/b0BM0al2crQBt6zovEtJfOc6/site/?key=statement", "Very Hard", false, [], ""),
    ]
  }
};

## Additional Reading

<resources>
  <resource source="CPH" title="18.4 - Merging Data Structures"></resource>
  <resource source="CF" title="Arpa - Sack (DSU on Tree)" url="blog/entry/44351"></resource>
  <resource source="CF" title="tuwuna - Explaining DSU on Trees" url="blog/entry/67696"></resource>
</resources>

## Merging Sets

<problems-list problems={metadata.problems.sam} />

Let's consider a tree rooted at node $1$, where each node has a color.

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 $O(N^2)$. 

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.

```cpp
if(a.size() < b.size()){ //for two sets a and b
  swap(a,b);
}
``` 

By merging the smaller set into the larger one, the runtime complexity becomes $O(N\log^2N)$.

### Proof

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 $O(N\log N)$.

<spoiler title="Full Code">

```cpp
#include <bits/stdc++.h>

using namespace std;

const int MX = 200005;

vector<int> adj[MX]; set<int> col[MX]; long long ans[MX];
void dfs(int v, int p){
    for(int e : adj[v]){
        if(e != p){
           dfs(e, v);
           if(col[v].size() < col[e].size()){
               swap(col[v], col[e]);
           }
           for(int a : col[e]){
               col[v].insert(a);
           }
           col[e].clear();
        }
    }
    ans[v] = col[v].size();
}
int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n; cin >> n;
    for(int i = 0; i < n; i++){
        int x; cin >> x;
        col[i].insert(x);
    }
    for(int i = 0; i < n - 1; i++){
        int u,v; cin >> u >> v;
        u--; v--;
        adj[u].push_back(v); adj[v].push_back(u);
    }
    dfs(0,-1);
    for(int i = 0; i < n; i++){
        cout << ans[i] << " ";
    }
}
```

</spoiler>

## Generalizing

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?).

## Problems

<problems-list problems={metadata.problems.general} />

<spoiler title="Solution to Promotion Counting">

```cpp
#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>

using namespace std;
using namespace __gnu_pbds;

template<class T> using Tree = tree<T,null_type,less<T>,rb_tree_tag,tree_order_statistics_node_update>;

const int MX = 1e5+5;
#define sz(x) (int)(x).size()

int N, a[MX], ind[MX], ans[MX], ret;
vector<int> child[MX];
Tree<int> d[MX];

void comb(int a, int b) {
  if (sz(d[a]) < sz(d[b])) d[a].swap(d[b]);
  for (int i: d[b]) d[a].insert(i);
}

void dfs(int x) {
  ind[x] = x; 
  for (int i: child[x]) {
    dfs(i);
    comb(x,i);
  }
  ans[x] = sz(d[x])-d[x].order_of_key(a[x]);
  d[x].insert(a[x]);
}

int main() {
  freopen("promote.in","r",stdin);
  freopen("promote.out","w",stdout);
  cin >> N; for (int i = 1; i <= N; ++i) cin >> a[i];
  for (int i = 2; i <= N; ++i) {
    int p; cin >> p;
    child[p].push_back(i);
  }
  dfs(1); 
  for (int i = 1; i <= N; ++i) cout << ans[i] << "\n";
}
```

</spoiler>

(also: same solution w/o indexed set)

## Faster Merging (Optional)

It's easy to merge two sets of sizes $n\ge m$ in $O(n+m)$ or $(m\log n)$ time, but sometimes $O\left(m\log 1+\frac{n}{m}\right)$ can be significantly better than both of these.

<resources>
  <resource source="CF" title="Splitting & Merging Segment Trees" url="https://codeforces.com/blog/entry/49446"></resource>
  <resource source="CF" title="Splitting & Merging BSTs" url="https://codeforces.com/blog/entry/67980"></resource>
</resources>

Requires knowledge of BSTs such as treaps or splay trees.

<problems-list problems={metadata.problems.treeRot} />