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C11XU

  • Title: C11XU - Bộ sưu tập đồng xu
  • Judge / source: VN SPOJ / VNOI
  • Original URL: https://vn.spoj.com/problems/C11XU/
  • Mirror URL: https://oj.vnoi.info/problem/c11xu
  • Secondary topics: Graphic matroid, Forest independence, Augmenting path
  • Difficulty: hard
  • Subtype: Choose at most one envelope per pair-group so the chosen envelope-edges stay acyclic
  • Status: solved
  • Solution file: c11xu.cpp

Why Practice This

This is a strong modeling problem.

The statement is about envelopes and parity, but the accepted reduction is:

  • each envelope becomes one edge between coin types
  • a bad subset is exactly a subset whose xor-parity is zero
  • in graph language, that means a cycle / even subgraph

So the chosen envelopes must form a forest.

Recognition Cue

Reach for this kind of reduction when:

  • the statement talks about parity or xor on chosen subsets
  • "bad subset" means some nonempty chosen family has total xor 0
  • each item naturally touches two labels or endpoints

The strongest smell is:

  • subset parity over pairs often becomes cycle-freeness in a graph model

Problem-Specific Transformation

The envelope story is rewritten as:

  • one envelope = one undirected edge between two coin types
  • a bad chosen subset = xor-sum 0

For edge-incidence vectors over GF(2), xor-sum 0 means every vertex has even degree, which is exactly the graph-language signal for a cycle / Eulerian subgraph.

So the original constraint becomes:

  • the chosen envelopes must stay acyclic

Then the extra "at most one from each consecutive pair" rule becomes a partition-side constraint on top of forest independence.

Core Idea

For an envelope containing coin types u and v:

  • if u = v, its parity vector is zero, so choosing it is immediately bad
  • otherwise it behaves like the incidence vector of an undirected edge (u, v)

A nonempty chosen subset is bad exactly when every type appears an even number of times inside that subset.

For edge-incidence vectors over GF(2), that means:

  • the chosen subset has xor-sum 0
  • equivalently, it contains a cycle / Eulerian subgraph

So valid chosen envelopes are exactly a forest in the type graph.

There is one more rule:

  • from each consecutive pair of envelopes, you may choose at most one

That is the intersection of:

  • a partition matroid on envelope-pairs
  • a graphic matroid on the type graph

With at most 1000 envelopes total, a small augmenting-path solver is fine.

Complexity

Let E = N / 2 be the number of envelopes.

The repo solution runs a cardinality matroid-intersection style augmentation in a size that is small enough for the judge:

  • time: good enough for E <= 1000
  • memory: O(E + M)

Pitfalls / Judge Notes

  • An envelope with two equal coin types can never be chosen.
  • “No bad subset” is stronger than “the whole chosen set is not bad”. You must forbid every nonempty chosen subset from having even parity on all types.
  • The right graph view is forest independence, not just DSU-greedy. A naive greedy choice can block a better later choice.

Reusable Pattern

  • Topic page: DSU
  • Practice ladder: DSU ladder
  • Starter template: dsu-basic.cpp
  • Notebook refresher: Data structures cheatsheet
  • Carry forward: state the set invariant first, then use DSU only for the merge logic it really supports.
  • This note adds: the extra problem-specific state layered on top of the basic union-find scaffold.

Solutions