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VOITSORT

  • Title: Cây hoán vị
  • Judge / source: VN SPOJ
  • Original URL: https://vn.spoj.com/problems/VOITSORT/
  • Secondary topics: Cartesian tree, Stack-sortable permutations, Next permutation
  • Difficulty: hard
  • Subtype: Count stack-sortable permutations in a short lexicographic interval
  • Source contest: VOI 2015 Day 2
  • Status: solved
  • Solution file: voitsort.cpp

Why Practice This

This is a strong permutation problem because the tree definition looks specialized, but the real key is recognizing a known permutation operator. After that, the second trick is to exploit the small bound k <= 10^6 < 10!.

Recognition Cue

This is a strong disguise problem for permutation operator + tiny moving suffix:

  • the statement defines a custom tree or traversal on permutations
  • the real property often matches a known permutation machine
  • the interval length k is large but still much smaller than n!
  • only a short lexicographic suffix can vary across the queried block

If a permutation problem gives both a structural definition and a short lexicographic window, try to identify the hidden operator first and only then optimize the enumeration.

Problem-Specific Transformation

Two rewrites unlock the whole solution:

  1. convert the TSort tree condition into the standard one-stack-sortable test
  2. use k <= 10^6 < 10! to freeze the first n - d positions and enumerate only the last d = min(n, 10) positions

So the problem stops being "count good permutations in a huge interval" and becomes:

  • inside each lexicographic block, test one fixed prefix once
  • iterate over a very small lexicographic suffix block
  • finish the stack simulation on that suffix

Core Idea

The binary tree T(pi) is the max-Cartesian tree of the permutation pi, and the required listing order is its postorder traversal.

A standard fact is:

  • postorder of the max-Cartesian tree of pi
  • equals the result of passing pi through one stack

So the TSort property means exactly:

  • pi is sortable by one stack into (1, 2, ..., n)

That is much easier to test. Simulate the usual stack-sorting process:

  1. read the permutation from left to right
  2. push each value
  3. while the stack top equals the next needed output, pop it

The permutation is TSort iff the whole process outputs 1, 2, ..., n.

Now for the counting part:

  • k <= 10^6 < 10!
  • therefore, among k consecutive permutations in lexicographic order, only the last 10 positions can vary freely
  • the interval crosses at most one boundary between blocks with fixed first n - 10 positions

So:

  1. split the interval into at most two blocks
  2. in each block, keep the prefix fixed
  3. enumerate only the suffix permutations of the last d = min(n, 10) positions
  4. test each full permutation by:
  5. simulating the fixed prefix once
  6. finishing the stack simulation on the small suffix

That gives O(n + k * d) with d <= 10.

Why Only The Last 10 Positions Matter

Fix the first n - d positions of a permutation. The remaining suffix has exactly d! lexicographic completions.

Choosing d = 10 works because:

  • 10! > 10^6
  • the interval length is only k <= 10^6

So a lexicographic interval of length k can cross at most one boundary between such suffix blocks. That is why the count splits into at most two manageable suffix-enumeration chunks.

This is a block-local statement: after splitting by block boundary, each chunk has one fixed prefix and one moving suffix. The entire original interval does not necessarily share one fixed prefix.

Complexity

  • block setup: O(n)
  • testing each candidate suffix: O(d)
  • total: O(n + k * d) with d <= 10

This is fast enough for n <= 1000, k <= 10^6.

Pitfalls / Judge Notes

  • The tree formulation is a disguise. The main equivalence is to one-stack sorting.
  • Do not iterate k times with full next_permutation on all n elements. That would be too slow.
  • The 10! observation is the whole optimization: only the suffix block changes significantly.
  • When the interval crosses the end of a suffix block, rebuild the next block once, then continue.

Reusable Pattern

Solutions