Disjoint Set Union
Authors: Benjamin Qi, Andrew Wang, Nathan Gong
Contributor: Michael Cao
The Disjoint Set Union (DSU) data structure, which allows you to add edges to a graph and test whether two vertices of the graph are connected.
Prerequisites
Focus Problem – try your best to solve this problem before continuing!
Resources
Resources | ||||
---|---|---|---|---|
CF EDU | video explanation + problems for DSU | |||
CSA | both optimizations, diagrams | |||
PAPS | both optimizations, no diagrams | |||
CPH | small to large, diagrams | |||
IUSACO | path compression, diagrams | |||
TC | diagrams |
Implementation
C++
Resources | ||||
---|---|---|---|---|
Benq (from KACTL) |
Below is an implementation of a DSU. It utilizes
small-to-large merging and path compression
to quickly perform union find. sizes[x]
stores the size of 's component,
and parents[x]
stores the parent of (equal to if it is the representative).
#include <bits/stdc++.h>using namespace std;class DisjointSets {private:vector<int> parents;vector<int> sizes;public:DisjointSets(int size) : parents(size), sizes(size, 1) {
Java
import java.util.*;public class DisjointSets {int[] parents;int[] sizes;public DisjointSets(int size) {parents = new int[size];sizes = new int[size];for (int i = 0; i < size; i++) {
Python
class DisjointSets:def __init__(self, size: int) -> None:self.parents = [i for i in range(size)]self.sizes = [1 for _ in range(size)]def find(self, x: int) -> int:""":return: the "representative" node in x's component"""if self.parents[x] == x:return xself.parents[x] = self.find(self.parents[x])
As the implementation is quite simple, you may prefer to use this in place of DFS for computing connected components.
Solution - Focus Problem
Without union find, we would have to represent the graph with an adjacency list and use flood fill to calculate connected components. This approach takes time, which is too slow, motivating us to use union find.
By representing the graph with the union find data structure that was just implemented above, we can use its methods to both unite vertices and check if two vertices and are in the same connected component using only amortized time.
This reduces the overall time complexity to , which is a substantial improvement and allows us to pass all test cases.
Implementation
Time Complexity:
C++
#include <bits/stdc++.h>using namespace std;Code Snippet: DSU (Click to expand)int main() {int node_num, query_num;cin >> node_num >> query_num;DisjointSets dsu(node_num);
Java
import java.io.*;import java.util.*;public class Main {public static void main(String[] args) {Kattio io = new Kattio();int size = io.nextInt();int queryNum = io.nextInt();DisjointSets dsu = new DisjointSets(size);
Python
Code Snippet: DSU (Click to expand)size, query_num = [int(i) for i in input().split()]dsu = DisjointSets(size)for _ in range(query_num):q_type, u, v = [int(i) for i in input().split()]if q_type == 0:dsu.unite(u, v)else:print(1 if dsu.connected(u, v) else 0)
Problems
Standard
You should already be familiar with the DFS / Binary Search solutions to "Wormhole Sort" and "Moocast."
Status | Source | Problem Name | Difficulty | Tags | |
---|---|---|---|---|---|
CSES | Easy | Show TagsDSU | |||
Gold | Easy | Show TagsDSU | |||
Gold | Easy | Show TagsDSU | |||
Silver | Easy | Show TagsDSU | |||
Gold | Easy | Show TagsDSU | |||
Old Silver | Easy | Show TagsDSU | |||
CSA | Normal | Show TagsDSU | |||
CF | Normal | Show TagsDSU |
Harder
Don't worry about solving these if this is the first time you've encountered DSU.
Status | Source | Problem Name | Difficulty | Tags | |
---|---|---|---|---|---|
CSES | Hard | Show TagsDSU, Merging | |||
Gold | Hard | Show TagsDSU, Merging, Sorted Set | |||
Old Gold | Hard | Show TagsDSU | |||
onlinejudge.org | Hard | Show TagsDSU | |||
Baltic OI | Very Hard | Show TagsDSU | |||
Gold | Very Hard | Show TagsDSU | |||
Platinum | Very Hard | Show TagsDSU |
Module Progress:
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