Overview

This problem asks us to color a tree map to minimize the total discomfort cost, which is defined as the sum of the difference between the color numbers of adjacent regions multiplied by the weight of the edge between them. Under the condition of “using at least 2 different colors”, the answer is the minimum value of the edge weights.

Analysis

First, the cost is defined as \(W_i \times |a - b|\) at each boundary. If there were no conditions, we could color all regions with the same color, which would make \(|a-b| = 0\) at all boundaries, resulting in a cost of \(0\). However, in this problem, we are not allowed to do this due to the constraint “use at least 2 different colors”.

Here, we make an important observation.

Observation 1: At least one boundary must connect different colors

Since the graph is connected and we must use at least 2 different colors, there must exist a pair of regions colored with different colors. Somewhere along the path connecting these two regions, the color must change, meaning there is at least one boundary where the colors on both ends are different. The cost of this boundary is: $\(W_i \times |a - b| \geq W_i \times 1 \geq (\text{minimum weight})\)\( (Since color numbers are integers, if they are different, their difference is at least \)1$).

Therefore, the total cost is always at least the “minimum weight”.

Observation 2: The minimum weight can be achieved exactly

In a tree, removing any single edge splits the graph into exactly two connected components. Thus, we can choose the edge \(e\) with the minimum weight, and color all regions in one of the components with color \(1\), and all regions in the other component with color \(2\).

Then: - Only the boundary corresponding to edge \(e\) connects regions with different colors, and its cost is \(W_e \times |1 - 2| = W_e\). - All other boundaries are within the same component, so both ends have the same color, resulting in a cost of \(0\).

Thus, the total cost is exactly \(W_e\) (the minimum weight).

Combining Observation 1 and Observation 2, we can see that the answer is the minimum value of the edge weights.

Algorithm

In fact, the structure of the tree itself does not affect the answer. We only need to find the minimum value of all edge weights \(W_i\) in the input.

You can solve this by simply updating the minimum value of \(W_i\) while reading the input. There is no need to construct or traverse the graph.

Complexity

• Time Complexity: \(O(N)\) (We only need to iterate through the \(M = N - 1\) edges once)
• Space Complexity: \(O(1)\) (Only a variable to keep track of the minimum value, excluding the space for reading input)

Implementation Points

• Since \(u_i\) and \(v_i\) are not needed for the answer, you can skip reading them and only keep track of the minimum value of \(W_i\).

• Since \(N\) can be as large as \(2 \times 10^5\), it is safe to read the input quickly. In the code, we can read the entire input at once using sys.stdin.buffer.read() and process it token by token.

• Since the number of colors \(K \geq 2\) and the number of regions \(N \geq 2\) are guaranteed, a valid coloring always exists, and the answer always exists (no special edge-case handling is required).

  Source Code

import sys

def main():
    data = sys.stdin.buffer.read().split()
    idx = 0
    n = int(data[idx]); idx += 1
    m = int(data[idx]); idx += 1
    k = int(data[idx]); idx += 1
    ans = None
    for _ in range(m):
        # u, v, w
        idx += 2  # skip u, v
        w = int(data[idx]); idx += 1
        if ans is None or w < ans:
            ans = w
    print(ans)

main()

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