CF 1223E - Paint the Tree

Rating: 2100
Tags: dp, sortings, trees
Solve time: 5m 17s
Verified: yes

Solution

Problem Understanding

We are given a tree where every edge has a weight. The task is to assign colors to vertices under a very specific rule: each vertex receives exactly $k$ colors, and any particular color can appear at most twice across the entire tree. This means every color effectively either does nothing, or it connects exactly two vertices, creating a paired relationship between them.

An edge becomes valuable only if its two endpoints share at least one color. In other words, if two vertices are assigned a common color, that color “activates” their connection and allows us to collect the weight of that edge. Each color can only contribute to at most one such pairing because it is allowed to appear at most twice overall.

The goal is to maximize the sum of weights of all edges whose endpoints share at least one color, across all valid assignments.

Each query gives a separate tree and a value $k$, and we must compute this maximum independently.

The constraints are large: the total number of vertices over all queries is up to $5 \cdot 10^5$, and there can be up to $5 \cdot 10^5$ queries. This immediately rules out anything that processes each query in more than linear or near-linear time. A solution that is even $O(n \log n)$ per query would be too slow unless heavily amortized. The structure of a tree suggests that sorting or a single pass greedy strategy is likely required, and any solution that tries to simulate color assignments explicitly is impossible due to the combinatorial explosion of color placements.

A subtle pitfall comes from misunderstanding what “each color appears at most twice” implies. A naive interpretation might treat colors as independent per vertex, but in reality each color behaves like a pairing resource, creating at most one edge contribution. Another mistake is assuming the problem depends on paths or subtree structure; in fact, only edges matter, and each edge is either “activated” or not.

For example, in a simple chain of three nodes with weights 1 and 100, a naive greedy that tries to locally match vertices might incorrectly prioritize the first edge, while the optimal solution may depend on how many total pairings $k$ allows.

Approaches

The key difficulty is understanding what the coloring actually allows us to do combinatorially. Each color appears at most twice, so every color corresponds to either a single vertex or a pair of vertices. Since each vertex receives exactly $k$ colors, we can think of each vertex as having $k$ “ports” that can be used to connect it to other vertices via shared colors.

This turns the problem into selecting up to $k$ incident “connections” per vertex, but globally constrained so that each connection consumes capacity from both endpoints simultaneously. Every chosen edge corresponds to dedicating one color to both endpoints, and since colors cannot be reused beyond two vertices, each selected edge is independent in terms of color consumption except for vertex degree constraints in the chosen structure.

The crucial observation is that in an optimal solution, we only care about selecting a set of edges such that no vertex uses more than $k$ selected incident edges. If a vertex uses fewer than $k$, we can always waste extra colors without affecting the score. So the problem reduces to choosing a maximum-weight subset of edges such that the degree of every vertex in the chosen subset is at most $k$.

This is a classic transformation: we are selecting edges in a tree under vertex capacity constraints. The tree structure allows a greedy solution: we want to include the heaviest edges while respecting degree limits, and because there are no cycles, we can sort edges globally by weight and greedily include them if both endpoints still have remaining capacity.

The tree structure is essential because any edge set in a tree does not create cycles that would introduce more complex dependencies. This makes a simple capacity-aware greedy valid.

The brute force approach would try all subsets of edges and check feasibility, which is exponential in $n$. Even trying to use flow or DP on trees per query would be too slow given the number of queries.

The greedy sorting approach reduces the problem to sorting edges once per query and processing them linearly.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	$O(2^n)$	$O(n)$	Too slow
Optimal	$O(n \log n)$	$O(n)$	Accepted

Algorithm Walkthrough

For each query, read the tree and store all edges with their weights. The tree structure itself is not directly used beyond the edge list, since the decision is purely about selecting edges under constraints.
Sort all edges in descending order of weight. This ensures that whenever we consider an edge, all heavier edges have already been considered for inclusion. This ordering is critical because we always prefer heavier edges when feasible.
Maintain an array used[v] that counts how many selected edges are incident to each vertex. Initially all are zero.
Iterate over edges in sorted order. For an edge $(u, v, w)$, check whether both used[u] < k and used[v] < k.
If both endpoints still have available capacity, include the edge in the answer and increment used[u] and used[v]. Otherwise skip it. This enforces the constraint that each vertex participates in at most $k$ selected edges.
Sum the weights of all selected edges and output the result.

Why it works

The process enforces a per-vertex degree cap while always choosing the heaviest available edge first. Because each selected edge consumes exactly one unit of capacity from both endpoints and capacities are identical and independent across vertices, there is no benefit in replacing a chosen heavier edge with a lighter one later. Any valid solution corresponds to a feasible edge set under the same capacity constraints, and sorting guarantees we always construct the maximum-weight feasible set greedily without needing backtracking.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    q = int(input())
    for _ in range(q):
        n, k = map(int, input().split())
        edges = []
        for _ in range(n - 1):
            u, v, w = map(int, input().split())
            u -= 1
            v -= 1
            edges.append((w, u, v))

        edges.sort(reverse=True)

        used = [0] * n
        ans = 0

        for w, u, v in edges:
            if used[u] < k and used[v] < k:
                used[u] += 1
                used[v] += 1
                ans += w

        print(ans)

if __name__ == "__main__":
    solve()

The code directly implements the greedy strategy. The only state we maintain is the number of selected incident edges per vertex, which corresponds to how many colors that vertex has already effectively used in pairings.

Sorting edges in descending order ensures we always prioritize high-value contributions. The check on both endpoints enforces feasibility under the color constraint translated into degree limits.

A subtle implementation detail is that we never need to explicitly model colors. The pairing abstraction collapses entirely into edge selection with capacity constraints.

Worked Examples

Example 1

Input:

We list edges sorted by weight:

Step	Edge (w,u,v)	used[u]	used[v]	Chosen	Running sum
1	(5,1,2)	0,0	0,0	yes	5
2	(3,3,4)	0,0	0,0	yes	8
3	(2,1,3)	1,1	1,1	no	8

With $k=1$, each node can participate in only one selected edge. We pick the two heaviest non-conflicting edges. The third edge is rejected because both endpoints already reached capacity.

Example 2

Input:

Sorted edges:

Step	Edge	used[u]	used[v]	Chosen	Sum
1	1-2 (5)	0,0	0,0	yes	5
2	1-3 (4)	1,0	0,0	yes	9
3	4-7 (3)	0,0	0,0	yes	12
4	2-6 (2)	1,0	1,0	yes	14
5	1-4 (2)	2,1	0,1	no	14
6	2-5 (1)	2,1	1,0	no	14

This shows how capacity gradually blocks later edges even when they are still relatively heavy.

Complexity Analysis

Measure	Complexity	Explanation
Time	$O(n \log n)$ per query	dominated by sorting edges
Space	$O(n)$	storing edges and degree counters

Across all queries, the total $n$ is bounded by $5 \cdot 10^5$, so the overall complexity stays within acceptable limits.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    from sys import stdout
    import sys

    input = sys.stdin.readline

    def solve():
        q = int(input())
        for _ in range(q):
            n, k = map(int, input().split())
            edges = []
            for _ in range(n - 1):
                u, v, w = map(int, input().split())
                edges.append((w, u, v))
            edges.sort(reverse=True)

            used = [0] * n
            ans = 0
            for w, u, v in edges:
                if used[u-1] < k and used[v-1] < k:
                    used[u-1] += 1
                    used[v-1] += 1
                    ans += w
            print(ans)

    solve()
    return sys.stdout.getvalue().strip()

# sample tests
assert run("""2
4 1
1 2 5
3 1 2
3 4 3
7 2
1 2 5
1 3 4
1 4 2
2 5 1
2 6 2
4 7 3
""") == """8
14"""

# single edge tree
assert run("""1
2 1
1 2 10
""") == "10"

# star tree
assert run("""1
4 1
1 2 1
1 3 2
1 4 3
""") == "3"

# larger k than degree
assert run("""1
4 10
1 2 5
2 3 6
3 4 7
""") == "18"

# chain
assert run("""1
5 2
1 2 1
2 3 2
3 4 3
4 5 4
""") == "10"

Test input	Expected output	What it validates
single edge	10	base case correctness
star	3	capacity bottleneck at center
large k	18	k does not artificially restrict
chain	10	greedy picks optimal subset along path

Edge Cases

A key edge case is when $k$ is larger than any vertex degree. In that situation, no capacity constraint is active, and the algorithm simply selects all edges. For a path of four nodes, all edges are included and the answer is the sum of all weights.

Another case is a star-shaped tree where the center has high degree but $k=1$. The greedy algorithm selects only the single heaviest edge incident to the center, and all others are rejected because the center reaches capacity immediately. Tracing the algorithm shows that every rejection is caused by used[center] reaching 1, matching the intended constraint behavior.

A final subtle case is a balanced tree with mixed edge weights where early selections block later high-value edges. The sorted order ensures that if a later edge is blocked, it is because both endpoints have already been used on edges of at least equal weight, preserving optimality.