CF 104686D - Deforestation
We are given a rooted tree where each node represents a physical segment of a large wooden structure. Each segment has a weight and may split into several child segments at its end.
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Solve time: 29s
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Solution
Problem Understanding
We are given a rooted tree where each node represents a physical segment of a large wooden structure. Each segment has a weight and may split into several child segments at its end. The root represents the trunk connected to the ground, and every node contributes its weight as part of the full tree.
We are allowed to cut the tree at arbitrary points along edges or within segments. After cutting, the tree breaks into connected components, and each component has a total weight equal to the sum of weights of its nodes. The goal is to partition the tree into the minimum number of pieces such that every resulting piece has total weight at most W.
The key output is the number of resulting connected components after performing an optimal set of cuts.
The input size is large in terms of structure. There are up to 10^5 segments, and total weight is bounded by 10^9. This immediately rules out any solution that recomputes subtree sums repeatedly in a naive way. A quadratic approach over nodes or repeated DFS recomputation would exceed limits.
The structure also hides a subtlety: branching is not special, it is just a tree, but cuts are not restricted to edges in a strict sense. However, because cutting inside an edge is equivalent to cutting at an edge boundary in terms of resulting connected components and weights, the problem reduces to deciding where to detach subtrees.
A few edge cases matter.
A single node whose weight exceeds W is impossible to pack into any piece without cutting inside it, but since cuts can be arbitrary, that node must itself be split into multiple pieces along its internal structure, effectively increasing the piece count even before considering children.
A star-shaped node with many children, each small, can still force many cuts if combined weights exceed W.
A long chain where cumulative weight barely exceeds W at many points is another case where greedy grouping decisions matter.
Approaches
A direct approach would be to compute subtree weights and try to greedily decide for each node how to group its children into valid chunks. One might attempt to simulate packing child subtrees into bins of capacity W, combining them arbitrarily.
The brute-force idea would be to treat each node as needing to distribute its children into groups such that each group plus the node weight stays within W, then recursively compute optimal grouping for each node. This quickly becomes combinatorial because each node with degree d may require partitioning its children into subsets, which is exponential in d in the worst case. With total nodes up to 10^5, this is infeasible.
The key observation is that this is fundamentally a tree knapsack-like problem, but with a strong monotonic structure: we never need to consider arbitrary groupings of children globally. Instead, optimality emerges from processing children in a DFS and accumulating their contributions greedily in a controlled way.
We root the tree and compute subtree sums. If a subtree exceeds W, it must be split internally, contributing extra pieces independently of the parent. Otherwise, it can potentially be merged upward into the parent’s component.
At each node, we are effectively deciding how many “carryable fragments” of its subtree can be merged upward without exceeding W. Any excess becomes a new piece.
This reduces the problem to a postorder traversal where each subtree returns a residual weight that can still be attached upward.
| Approach | Time Complexity | Space Complexity | Verdict |
|---|---|---|---|
| Naive subset grouping per node | Exponential | O(n) | Too slow |
| Postorder greedy aggregation | O(n) | O(n) | Accepted |
Algorithm Walkthrough
We treat the tree as rooted at the trunk and compute a DFS from the root.
- Perform a postorder traversal so that all child subtrees are processed before their parent. This ensures we always know the “residual carryable weight” of each child before