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321C - Ciel the Commander

  • Title: Ciel the Commander
  • Judge / source: Codeforces
  • Original URL: https://codeforces.com/problemset/problem/321/C
  • Secondary topics: Centroid decomposition, Centroid tree, Balanced recursive labeling
  • Difficulty: hard
  • Subtype: Build the centroid decomposition tree itself and label nodes by centroid depth
  • Status: solved
  • Solution file: cielthecommander.cpp

Why Practice This

This is the cleanest in-repo anchor for trusting the centroid decomposition structure itself.

The problem does not hide behind a second data structure or a later query trick.

It asks for a labeling whose proof is exactly the decomposition:

  • choose one balanced centroid
  • give it the highest available rank
  • recurse on the remaining components

So the real lesson is:

  • how to find a centroid repeatedly
  • how the centroid tree is built
  • why nodes with the same centroid depth automatically satisfy the path rule

Recognition Cue

Reach for centroid decomposition when:

  • the tree is static
  • the statement wants a balanced recursive partition
  • or every solution argument sounds like “pick one central node, solve paths through it, recurse on the rest”

The strong smell here is:

  • if two nodes share the same rank, some higher-ranked node must lie on their path

That is exactly what centroid levels provide.

Problem-Specific Transformation

Let the rank of a node be determined by its depth in the centroid tree:

  • root centroid -> A
  • its centroid-tree children -> B
  • next level -> C
  • and so on

Then if two nodes x and y receive the same rank, they lie in different child components of some higher centroid chosen earlier.

So the unique path between x and y passes through that earlier centroid, which has a strictly higher rank.

That matches the statement directly.

Core Idea

Build the centroid decomposition tree of the original tree.

For each recursive component:

  1. find its centroid c
  2. label c with the current letter
  3. remove c
  4. recurse on every remaining component with the next letter

Why is the labeling valid?

Because equal letters correspond to equal centroid-tree depth.

Two nodes at the same centroid depth were separated earlier by some ancestor centroid with a smaller depth, hence a higher rank.

Therefore the path between them contains a higher-ranked node.

This is one of the rare problems where the centroid decomposition is not just a helper.

It is the whole proof.

Complexity

  • each decomposition level touches every live node once
  • centroid-tree depth is O(log n)
  • total time: O(n log n)
  • memory: O(n)

This is easily fast enough for n <= 10^5.

Pitfalls / Judge Notes

  • The rank depends on centroid depth, not ordinary tree depth.
  • Do not confuse the centroid tree with the original tree.
  • Using a removed[] array is safer than erasing edges from adjacency lists.
  • In this constraint range, centroid depth stays well below 26, so a valid labeling exists.

Reusable Pattern

Solutions