### Jon.Snow's blog

By Jon.Snow, 23 months ago,

Hello Codeforces,

CSES is a nice collection of classical CP problems which encourages you to learn a lot of basic and advanced concepts. Having editorials would help people to not get stuck on a problem for long. Here are my solutions to the tree section of the problem-set. Looking forward to read corrections/feedback/better solutions in the comment section. Happy coding!

## Subordinates

This can be solved using basic DP approach in linear time.
$subordinates[u] = \sum\limits_{v:children[u]}^{} subordinates[v]$ where v is a child of u

Code

## Tree Matching

This problem can be reduced to maximum bipartite matching problem but first we need to split the tree into a bipartite graph. One way to do that is to put all nodes at even depth in the first group and odd ones in the other group. Such a splitting will make sure that we do not have any edges within the same group.

Next, we can use Hopkraft Karp algorithm to find the maximum matching in the bipartite graph created.

Code

## Tree Diameter

This is a classical problem having multiple solutions.

One easy to implement solution is using 2 Breadth First Searches (BFS). Start a BFS with a random node and store the last node encountered before search ends. This last node will definitely be one of the ends of the diameter (Why?). Now run a second BFS from this node and you will end on the other end of the diameter.

Code

## Tree Distances I

For each node, we want to calculate maximum distance to another node. Previously we saw that that if we start a BFS from any node, we end up on either of the diametric end. We can use this fact to efficiently compute the answer. Let's calculate distances of each node from both the ends of the diameter. Then maximum distance of each node can be calculated as:

max_distance[u] = max(distance_from_diametric_end1[u], distance_from_diametric_end2[u])

Code

## Tree Distances II

We can solve this problem using in-out DP on trees.
$in[u]:$ sum of distances from u to each node in subtree rooted at u
$out[u]:$ sum of distances from u to each node excluding the subtree rooted at u

Now, ans[u] = in[u] + out[u]

Calculating $in[u]$ is quite straightforward. Suppose we want to calculate the sum of distances starting at node u and ending at any node in subtree rooted at v. We can use the pre-calculated value for v and separately add the contribution created by edge $u\rightarrow v$. This extra quantity will be the subtree size of u. Then we can repeat the process for each child of u.

$in[u] = \sum\limits{}{}in[v] + sub[v]$ where v is a child of u

To calculate $out[u]$, first let's calculate the contribution of parent of u. We can use out[par] and add the difference created by the edge $u \rightarrow par_u$ which is n-sub[par]+1. Next, we add the contribution of each sibling of u: in[sib] + sub[sib]*2.
Finally we have the following formula:

$out[u] = out[par] + (n-sub[par] + 1) + (\sum\limits_{sib}{} in[sib] + sub[sib]*2) - (in[u] + sub[u]*2)$

Code

## Company Queries I

We can solve this problem using binary-lifting technique for finding ancestors in a tree.

For each node x given in a query, we just need to find the $k^{th}$ ancestor of a given node. We can initialise ans = x and keep lifting our ans for every $i^{th}$ bit set in k: ans = jump[i][ans] where $jump[i][j]$ holds $i^{th}$ ancestor of node $j$

Code

## Company Queries II

This is the classical problem of finding Lowest Common Ancestor which can be solved in multiple ways. One of the common ways is to use binary-lifting which requires $O(nlog(n))$ preprocessing and $O(logn)$ per query.

Code

## Distance Queries

Distance between node u and v can be calculated as $depth[u] + depth[v] - 2*depth[LCA(u,v)]$.
LCA of (u,v) can be calculated using binary-lifting approach in $O(logn)$ per query.

Code

## Counting Paths

This problem can be solved using prefix sum on trees.

For every given path $u \rightarrow v$, we do following changes to the prefix array.

prefix[u]++
prefix[v]++
prefix[lca]--
prefix[parent[lca]]--


Next, we run a subtree summation over entire tree which means every node now holds the number of paths that node is a part of. This method is analogous to range update and point query in arrays, when all updates are preceded by queries.
$prefix[u] = \sum\limits_{v:children[u]}^{} prefix[v]$

Code

## Subtree Queries

This problem can be solved by flattening the tree to an array and then building a fenwick tree over flattened array.

Once reduced to an array, the problem becomes same as point update and range query. We can flatten the tree by pushing nodes to the array in the order of visiting them during a DFS. This ensures the entire subtree of a particular node forms a contiguous subarray in the resultant flattened array. The range of indices corresponding to subtree of a node can also be pre-calculated using a timer in DFS.

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## Path Queries

This problem can be solved using Heavy-Light decomposition of trees.

First, we decompose the tree into chains using heavy-light scheme and then build a segment tree over each chain. A path from node u to v can seen as concatenation of these chains and each query can be answered by querying the segment trees corresponding to each of these chains on the path. Since heavy-light scheme ensures there can be at most $O(logn)$ chains, each query can be answered in $O(log^{2}n)$ time.

Similarly, each update can be performed in $O(logn)$ time as it requires update on a single chain (single segment tree) corresponding to the given node.

Code

## Distinct Colors

This problem can be solved using Mo's algorithm on trees and can be reduced to this classical SPOJ Problem

Flatten the tree to an array using the same technique mentioned in solution of Subtree Queries. Then the subtree of each node will correspond to a contiguous subarray of flattened array. We can then use Mo's algorithm to answer each query in $O(\sqrt{n})$ time with $O(n\sqrt{n})$ preprocessing.

Code
Alternate solution using small-to-large technique

## End Notes

• +84

 » 22 months ago, # |   0 It's been 4weeks and still, no one reacted?
•  » » 22 months ago, # ^ | ← Rev. 2 →   +3 This probably means he started working on the blog 4 weeks ago and not that he posted it 4 weeks ago.How long have you been using CF, smh xD
•  » » » 22 months ago, # ^ |   0 Lol
 » 22 months ago, # |   0 Good job on the editorial! you might like my DP solution(check code) for tree matching.
•  » » 22 months ago, # ^ |   0 Thanks for mentioning! Seems like maximum matching was on overkill.
 » 22 months ago, # |   +6 Thanks for this editorial, might motivate many people like me to practice tree problems!
 » 22 months ago, # |   0 Can someone please give the link for problemset.
•  » » 22 months ago, # ^ |   +16 https://cses.fi/problemset/list/Jon.Snow Maybe add it to the blog post?
•  » » » 22 months ago, # ^ |   0 Thank you!
•  » » » 22 months ago, # ^ |   0 Yes, added to the post.
 » 22 months ago, # |   +3 Post is itself a good one, just a add on, read this awesome blog by adamant on heavy- light decomposition, using this we can solve both path queries as well as subtree queries simultaneously.
 » 22 months ago, # |   +3 There are some problems that can be done better :) Tree MatchingDP$dp(u, 0) =$ maximum number of offices in subtree of $u$, suppose $u$ is not included in matching$dp(u, 1) =$ maximum number of offices in subtree of $u$, suppose $u$ is included in matching GreedyJust take any leaf in the tree when possible Path QueriesThis one is special, since all path queries are from root to node.Let's calculate the answer $ans[u]$ for every nodes $u$ at first.When the value of a node $u$ changes, for which nodes do its answer changes? AnswerThe subtree of $u$ :)So we can update answer of the entire subtree of $u$ by flattening the tree. Now the problem is literally "Subtree Queries". Distinct ColorsThis is an application of the small-to-large technique.For each node of the tree, we maintain a set for all of its colors. When we need to merge a "parent" set with a "child" set, we simply merge the smaller child into bigger child, instead of otherwise. Codeset s; void dfs(int u, int prev) { set t; t.insert(col[u]); for (auto v: adj[u]) { if (v != prev) { dfs(v, u); if (s.size() > t.size()) swap(t, s); for (int x: s) t.insert(x); } } ans[u] = t.size(); s.swap(t); } And simply call dfs for any starting node you like.
•  » » 22 months ago, # ^ |   0 Nice observations! Will definitely try all these approaches.
•  » » » 9 months ago, # ^ |   0 If you implemented these, could you please add them to the editorial above. Novice people like me learn through implementation off problems much more than plain comments.
•  » » 11 months ago, # ^ |   0 Tree Matching :Greedy — sort all the edges according to the number of subordinates then check and count whether the nodes in the edge is taken or not :)
 » 22 months ago, # |   0 What is the full form of CSES?
•  » » 8 days ago, # ^ |   0 code submission evolution system
 » 22 months ago, # |   +8 You know many thing Jon.Snow.
•  » » 22 months ago, # ^ |   +3 Guarding the wall gets too boring
 » 22 months ago, # |   0 Thanks for the mention :)
 » 22 months ago, # |   +3 Thanks for this! You can also add Graph editorials for CSES problem set to the list. They are not yet complete though
 » 22 months ago, # |   +8 For Tree Distances Imax_distance[u] = max(distance_from_diametric_end1[u], distance_from_diametric_end2[u])Is there any proof for this?
•  » » 22 months ago, # ^ |   +6 First, suppose that $u$ is lying on diameter (path from $end1$ to $end2$). Now lets prove it by contradiction: suppose we can do (strictly) better. Without lose of generality, lets say that end1 is in the same "side" of the tree where is the node with whome we can do better (lets name that node $v$). So since $v$ is on the same side as end1, and you need to pass throw $u$ to get to that side from $end2$, then distance from $end2$ to $v$ can be shown as sum of distance (from $end2$ to $u$) + (from $u$ to $v$). And we assumed that distance from $u$ to $end1$¸is shorter then from $u$ to $v$. Then distance from end2 to $v$ is (strictly) greater than distance from end2 to end1. (but end1 to end2 is diameter — longest distance possible!)Simular can be shown if $u$ is not on diameter. Try working it out yourself! Good luck :)
 » 22 months ago, # |   +1 Thanks for the editorial bro.
 » 22 months ago, # |   +1 Nice editorial man! Please keep adding stuffs like this. They are really helpful atleast for newbies and beginnners!
 » 21 month(s) ago, # |   0 For Tree Diemeter, it can also be solved with 1 dfs instead of 2 dfs / 2 bfs. codevoid dfs(int u, int p){ EACH(v, g[u]){ if(v^p){ dfs(v,u); MAX(ans,d[u]+d[v]+1); MAX(d[u],d[v]+1); } } } dfs(0,-1); 
 » 20 months ago, # |   0 Can Distinct Colors be solved using Segment Tree? I tried but it gives TLE as merging two sets is taking linear time How do i optimise it?
•  » » 20 months ago, # ^ | ← Rev. 2 →   +1 Transform the problem into a range query problem on an array using dfs order (Let this array of colors be A)Let Prev[i] be the maximum index `j
 » 13 months ago, # |   0 In tree Distances (II) what does the sub[v] means ?
 » 11 months ago, # |   +3 Path Queries 1 , could be solved using Tree flattening and Segment tree . The basic idea is that change in value of node 's' changes only the values of all nodes in its subtree , so we can use range update , point query segment tree to solve it in O(NlogN) AC solution
 » 9 months ago, # |   0 How to solve the fixed length path 1 and 2 ? Any editorial for this one ?