CF 1085D - Minimum Diameter Tree

CF 1085D - Minimum Diameter Tree

Rating: 1700
Tags: constructive algorithms, implementation, trees
Solve time: 2m 20s
Verified: no

Solution

Problem Understanding

We are given a tree and a fixed total amount of “weight budget” $s$. Every edge must be assigned a non-negative real weight, and the sum over all edges must equal exactly $s$. Once weights are assigned, the distance between two vertices is the sum of edge weights along their unique path in the tree, and the diameter is the maximum such distance over all pairs of vertices.

The task is to distribute the total weight across the edges so that the longest root-to-leaf style path anywhere in the tree becomes as small as possible.

The key difficulty is that we are not choosing a path but distributing mass globally across a tree structure. Any edge we make heavy increases all paths that pass through it, so the optimal configuration depends entirely on how many “important directions” the tree has, meaning how many branches compete to form long paths.

The constraints allow up to $10^5$ nodes, so any solution must be essentially linear or near-linear in the size of the tree. A quadratic or even $O(n \log^2 n)$ construction would be risky if it involves repeated global recomputation. This strongly suggests that the answer depends only on a small structural parameter of the tree rather than on any detailed configuration of weights.

A few edge cases expose what is non-trivial.

If the tree is a simple path, there is no branching. Any distribution of weights still creates a single path whose total weight is always $s$, so the diameter is forced to be $s$. A naive attempt to “spread weight” cannot reduce it.

If the tree is a star, all paths go through the center. It becomes possible to split the total weight across many edges, so no single path accumulates much weight. For example, with $n=4$ and a star centered at 1, assigning equal weights $s/3$ yields diameter $2s/3$. This shows that branching reduces the effective diameter.

If a method incorrectly assumes that the best structure is always a path or always a star-like distribution, it fails on the opposite case. The correct solution must adapt to the number of leaves.

Approaches

A brute-force viewpoint starts by imagining we guess all edge weights subject to the sum constraint and compute the diameter by running DFS between all pairs of vertices. Even if we only try to optimize weights in a continuous sense, we would still need to reason over infinitely many assignments, and even discretizing would explode combinatorially. Any attempt to simulate candidate distributions leads to at least exponential complexity in the number of edges.

The key observation is that only the number of leaves matters, not the exact shape of internal branching. Once a tree has $k$ leaves, any path contributing to the diameter can be thought of as going from one leaf to another through some internal structure. The optimal strategy tries to “push” weight toward the leaves evenly, so no pair of leaves becomes too far apart.

The fundamental structural idea is to orient the tree around a central point so that all leaves are connected through a balanced center. Then the worst path is determined by how many leaf-branches are forced to share the total weight. If there are many leaves, the weight must be split more finely, reducing the maximum pairwise distance. If there are only two leaves (a path), no such splitting helps.

This leads to a simple characterization: if there are $k$ leaves, the optimal diameter is proportional to $2s/k$. Intuitively, each leaf contributes a “budget share” of approximately $s/k$, and the longest path between two leaves uses two such contributions.

This collapses the problem from a global optimization over tree metrics to a single counting problem.

Algorithm Walkthrough

1. Count the number of leaves in the tree. A leaf is a vertex with degree 1. This step captures the number of independent endpoints that must “share” the total weight budget.
2. If the tree has exactly two leaves, return $s$. This corresponds to a path graph, where all weight lies on a single chain and cannot be redistributed across branches.
3. Otherwise, compute $k$, the number of leaves.
4. Compute the answer as $\frac{2s}{k}$. This represents the worst-case distance between two leaves when the total weight is evenly distributed across all leaf branches.
5. Output this value as a floating-point number.

The reasoning behind step 4 is that every leaf-to-leaf path must pass through the internal core of the tree. If each leaf is assigned an equal share of the total weight, then the distance from a leaf to the center is effectively $s/k$, and a diameter path consists of two such segments.

Why it works

Any tree can be decomposed into a core where internal vertices distribute flow between leaves. Since every leaf contributes