Summary

We are coloring each vertex of a tree structure with \(K\) types of colors. Under the condition that at least two different colors must be used, the problem asks us to minimize the sum of “difference in color IDs \(\times\) edge weight” between adjacent vertices.

Observation

To solve this problem, we focus on the properties of a tree and the lower bound of the cost.

First, coloring all vertices with the same color is forbidden. That is, we must use at least two different colors. Using at least two colors means that in at least one edge of the tree, the colors of the two endpoints must be different.

When adjacent vertices have different colors, the cost for that edge is \(W_i \times |a - b|\). Since \(a \neq b\), we have \(|a - b| \ge 1\). Therefore, the cost of an edge with different colors is at least \(W_i \times 1 = W_i\). Since the total cost is the sum of the costs of each edge, it must be at least the weight of some edge. Thus, we can see that the total cost is at least the minimum weight of the edges in the tree, \(\min(W_i)\).

Next, we consider whether this lower bound \(\min(W_i)\) can actually be achieved. If we choose any single edge from the tree and remove it, the tree is split into two subtrees (since a tree is connected and has no cycles). We choose the edge with the minimum weight and split the tree into two parts using this edge. Then, we color all vertices belonging to one subtree with “Color 1” and all vertices belonging to the other subtree with “Color 2”. With this coloring, only the endpoints of the edge used for splitting will have different colors (Color 1 and Color 2), and its cost will be \(W_{\min} \times |1 - 2| = W_{\min}\). All other edges connect vertices of the same color, so their cost is \(0\). Since it is guaranteed that \(K \ge 2\), this coloring is always possible, and the total cost will be exactly \(W_{\min}\).

Therefore, the answer is the minimum value of all edge weights \(W_i\).

Algorithm

1. Read all edge weights \(W_i\) from the input.
2. Find the minimum value among them.
3. Output the minimum value.

Complexity

• Time Complexity: \(O(N)\) (since we examine each edge weight once)
• Space Complexity: \(O(1)\) (processing sequentially without storing all inputs, or \(O(N)\) which is well within the limit even if we read the entire input into memory)

Implementation Details

To process a large amount of input quickly in Python, we can use sys.stdin.buffer.read().split() to read the entire input at once and split it by whitespace. In the input data, the edge weight \(W_i\) corresponds to the third element of each line. Since the beginning of the input data consists of three elements N M K, the weights \(W_i\) are located at indices 5, 8, 11, ... of the data list. By scanning only these values and updating the minimum, we can find the solution very simply and quickly.

Source Code

import sys

def solve():
    data = sys.stdin.buffer.read().split()
    if not data:
        return
    
    N = int(data[0])
    M = int(data[1])
    # K = int(data[2]) # 使用しない
    
    min_W = 10**9
    # 各行の3番目の要素が W_i にあたる
    # dataのインデックスは 5, 8, 11, ...
    # 開始インデックスは 5
    for i in range(5, len(data), 3):
        w = int(data[i])
        if w < min_W:
            min_W = w
            
    print(min_W)

if __name__ == "__main__":
    solve()

This editorial was generated by or-glm5.2-high.