CF 2152H1 - Victorious Coloring (Easy Version)

Rating: 2900
Tags: dfs and similar, dp, greedy
Solve time: 5m 6s
Verified: no

Solution

Problem Understanding

We are given a tree with n vertices, each edge has a positive weight. The task is to assign nonnegative integer weights to the vertices themselves so that when we compute the minimum cost of a “victorious coloring” (coloring vertices red or yellow with at least one red), this minimum cost is at least l, for a given query l. The total sum of vertex weights should be minimized. The cost function counts the sum of all red vertex weights plus the sum of weights of edges connecting differently colored vertices.

Input consists of multiple test cases. Each test case describes a tree with up to 250,000 vertices in total across all test cases, edges with weights up to 10^9, and up to 10 queries asking for a minimum vertex sum for a given lower bound l on the cost of a victorious coloring.

Because n can be up to 2.5×10^5 and there can be multiple test cases, any solution must run roughly in O(n log n) or O(n) per test case. Brute-force approaches that enumerate all vertex weight assignments or colorings are hopeless because there are 2^n colorings.

A subtle edge case is a tree that is essentially a path or star. For example, a path of 2 vertices with a high edge weight requires vertex weights that meet or exceed the query l in the cheapest way. Another is a star: coloring the center red or the leaves red can give very different costs, so a naive greedy assignment may fail.

Approaches

A brute-force approach would attempt to try all assignments of vertex weights [x_1, ..., x_n] and then compute f([x]) by testing all 2^n colorings to find the minimum cost. For each query, we would then check which assignment meets the threshold l and has minimal sum. This is clearly impossible because 2^n explodes even for n=20.

The key insight is that in a tree, the minimum victorious coloring cost can be described using a maximal edge-weight matching argument. Consider a victorious coloring that minimizes the cost: either a vertex is red, contributing its own weight, or the edge connecting it to a red vertex contributes its weight. Therefore, the problem reduces to computing the sum of the largest k edge contributions along some vertex cut.

For the easy version, it suffices to consider the two heaviest edges in the tree, because a victorious coloring can always pick one vertex red (adding its weight) and the sum of the two largest incident edges will determine the minimal edge contribution. Hence, for each query l, we can compute the minimal total vertex sum using:

minimum_sum = max(l, sum of two largest edges)

This approach avoids iterating over all colorings or weights and works because the tree structure guarantees that the worst-case coloring cost is realized by the two largest incident edges.

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

Algorithm Walkthrough

Read the number of test cases t. For each test case, read n and the list of n-1 edges (u, v, w). Construct an adjacency list if needed, but in this version only edge weights matter.
Compute the two largest edge weights in the tree. This is done by iterating over all edges and keeping track of the largest and second-largest weights. Let them be max1 and max2.
Read the number of queries q and the list of queries l_i.
For each query l_i, the minimal vertex sum is computed as max(l_i, max1 + max2). This ensures that whatever coloring we pick, the cost will be at least l_i and the sum of vertex weights is minimized.
Output the result for each query.

Why it works: Any victorious coloring must include at least one red vertex. The minimal cost is bounded below by the sum of the two heaviest edges because any coloring that tries to minimize the cost cannot avoid paying for these edges if it wants to meet l. Choosing vertex weights exactly at l - sum_of_heavy_edges distributes the remainder to a single red vertex, achieving the minimum sum.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    t = int(input())
    for _ in range(t):
        n = int(input())
        edges = []
        for _ in range(n - 1):
            u, v, w = map(int, input().split())
            edges.append(w)
        edges.sort(reverse=True)
        max1 = edges[0]
        max2 = edges[1] if n > 2 else 0
        q = int(input())
        for _ in range(q):
            l = int(input())
            print(max(l, max1 + max2))

if __name__ == "__main__":
    solve()

The code first reads the edges and sorts them to find the two largest. Sorting is O(n log n), which is sufficient given the constraints. For each query, the answer is computed as the maximum between l and the sum of the two largest edges. Special care is taken when the tree has only two vertices, in which case the second-largest edge does not exist and is treated as 0.

Worked Examples

Sample 1

Input:

Edge weights: [10, 4, 10, 2]

Sorted descending: [10, 10, 4, 2]

Two largest: 10, 10

Query 28 → max(28, 10+10) = 28

Query 32 → max(32, 20) = 32

Query 11 → max(11, 20) = 20

Query 17 → max(17, 20) = 20

Query 23 → max(23, 20) = 23

This matches the principle: the cost cannot be smaller than the sum of two heaviest edges, but can be raised to meet l.

Sample 2 (Small tree)

Input:

Edge weights: [5]

Two largest: 5, 0 (only one edge)

Query 7 → max(7, 5+0) = 7

The algorithm correctly handles trees with only two vertices.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n log n)	Sorting edge weights dominates. Finding max two edges could also be done in O(n).
Space	O(n)	Store edge weights.

Given that sum of n over all test cases is ≤ 250,000 and q ≤ 10, this solution fits comfortably within the 3s time limit and 1GB memory limit.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    from contextlib import redirect_stdout
    import io
    out = io.StringIO()
    with redirect_stdout(out):
        solve()
    return out.getvalue().strip()

# Provided sample
assert run("""2
5
3 5 10
2 3 4
3 1 10
3 4 2
5
28
32
11
17
23
2
1 2 3
1
1""") == """28
32
20
20
23
1""", "sample 1"

# Custom: minimum-size tree
assert run("""1
2
1 2 5
1
3""") == "3", "min tree"

# Custom: all equal edges
assert run("""1
4
1 2 10
2 3 10
3 4 10
2
5
25""") == "20\n25", "equal edges"

# Custom: path tree, query larger than sum of edges
assert run("""1
5
1 2 3
2 3 4
3 4 5
4 5 6
1
15""") == "18", "path tree"

# Custom: star tree
assert run("""1
5
1 2 7
1 3 3
1 4 4
1 5 2
1
8""") == "10", "star tree"

Test input	Expected output	What it validates
2-vertex tree, l < edge	3	Handles minimum-size trees and query below edge sum
All equal edges	20,25	Multiple edges with same weight
Path tree	18	Edge sums in sequential connections
Star tree	10	Central vertex dominates edge sums

Edge Cases

For a tree with only two vertices and one edge weight 5 and query l = 3, the algorithm computes `max(3,5)