Skip to content

VOSTRIP

Why Practice This

This is a great tree problem because the final solution is much simpler than it first looks. Instead of building the paths directly, you only need to understand what must happen locally at each vertex.

Recognition Cue

Reach for local degree-balancing on a tree when:

  • the input gives required path usages on edges
  • the final object is a set of simple paths
  • direct construction of all paths looks messy
  • each vertex only needs to know how many incident path copies can be paired internally

The strongest smell is:

  • "minimum number of paths realizing edge-usage counts on a tree"

That often means count endpoints first and divide by two, instead of constructing paths directly.

Problem-Specific Transformation

The statement is rewritten as:

  • each edge with count c contributes c path copies through that incidence
  • at one vertex, incident copies are paired whenever they continue through different edges

So the hard global path-cover problem becomes:

  • local pairing and leftover counting at each vertex

Those leftovers are exactly path endpoints, which makes the final answer one global endpoint count.

Core Idea

Each tour is a simple path on the tree. Every time a path uses an edge, that edge count decreases by 1.

Think locally at a vertex u. Suppose the incident edges of u have required usage counts:

c1, c2, ..., ck

Let:

  • S = c1 + c2 + ... + ck
  • M = max(ci)

Every path copy that reaches u through one incident edge can:

  • continue through a different incident edge, or
  • stop at u, which creates one endpoint there

So at u, we want to pair as many incident edge-copies as possible, but we are never allowed to pair two copies from the same edge together.

That means the minimum number of unpaired copies left at u is:

need(u) = max(S mod 2, 2 * M - S)

Why:

  • S mod 2 handles parity: if S is odd, at least one copy must stay unmatched
  • 2 * M - S handles domination: if one edge contributes more copies than all other incident edges combined, the excess cannot be paired away

Those unmatched copies are exactly path endpoints at u.

So the minimum total number of tours is:

answer = (sum of need(u) over all vertices) / 2

because every path has exactly two endpoints.

Complexity

  • time: O(N)
  • memory: O(N)

This is important because the largest tests go up to N <= 10^6, while the edge counts can be as large as 10^12.

Pitfalls / Judge Notes

  • You do not need adjacency lists or DFS. The answer depends only on:
  • the sum of incident weights at each vertex
  • the maximum incident weight at each vertex
  • Use long long.
  • The official statement comments mention that the test data avoids pathological looping ambiguities from the story interpretation. The endpoint formula is still the correct accepted model.

Reusable Pattern

  • Topic page: Tree DP
  • Practice ladder: Tree DP ladder
  • Starter template: tree-dp-subtree.cpp
  • Notebook refresher: Graph cheatsheet
  • Carry forward: root the tree and make every subtree contribution local before combining children.
  • This note adds: the exact aggregation rule and path interpretation needed here.

Solutions