Overview

This problem asks us to color a tree-structured map to minimize the sum of costs, where the cost of each boundary is the absolute difference between the color IDs of adjacent regions multiplied by the weight of the boundary. Under the constraint that “at least 2 different colors must be used,” the answer is the minimum weight among all edges.

Analysis

First, let’s look at the definition of the cost: \(W_i \times |a - b|\). Since \(W_i \geq 1\) and \(|a-b| \geq 0\), the cost of each boundary is always at least \(0\).

If there were no constraint on the number of colors, we could color all regions with the same color ID, making \(|a-b| = 0\) for all boundaries, which results in a total cost of \(0\). However, because of the constraint “at least 2 different colors must be used” in this problem, we cannot do this.

Therefore, we need to find a way to “use at least 2 different colors while avoiding unnecessary costs as much as possible.”

Key Observation: Since the graph is connected (a tree), if we use at least 2 different colors, there must be at least one boundary where the colors of the two adjacent regions are different.

This is because if the colors on both sides of every boundary were the same, the color would “propagate” across the entire connected graph, resulting in all regions having the same color. This contradicts the constraint of “using at least 2 different colors.”

Therefore, there is at least one boundary with different color IDs, and the cost of this boundary is at least:

\[W_i \times |a - b| \geq W_i \times 1 = W_i \geq \min_j W_j\]

Since the costs of all other boundaries are at least \(0\), the total cost is always at least \(\min_j W_j\).

Algorithm

This lower bound of \(\min_j W_j\) can actually be achieved. Specifically, we can color the regions as follows:

1. Choose one edge \(e\) with the minimum weight.
2. Removing edge \(e\) from the tree splits it into two connected components.
3. Color all regions in one component with Color 1, and all regions in the other component with Color 2.

With this coloring scheme:

• The endpoints of edge \(e\) have Color 1 and Color 2, so its cost is \(W_e \times |1 - 2| = W_e\).
• For all other boundaries, both endpoints belong to the same component, so they have the same color ID, resulting in a cost of \(0\).

Thus, the total cost is exactly \(W_e = \min_j W_j\). Since \(K \geq 2\), we can always use Color 1 and Color 2.

Therefore, the answer is the minimum weight among the input edges.

In terms of implementation, there is no need to even store the tree structure (adjacency list). You only need to find the minimum weight while reading the edges.

Complexity

• Time Complexity: \(O(M)\) (just reading each edge once)
• Space Complexity: \(O(1)\) (only need to keep track of the minimum weight)

Implementation Points

• Since the weights are \(W_i \leq 10^5\) and the answer is the weight of a single edge, it would fit in a standard int. However, we can use long long just to be safe.

• Initialize the minimum value to LLONG_MAX and update it by comparing it with the weight of each edge.

• The values of \(u_i\) and \(v_i\) themselves are not needed for the answer, so you can simply ignore them while reading the input (no need to build the tree).

  Source Code

#include <bits/stdc++.h>
using namespace std;

int main(){
    int N, M;
    long long K;
    scanf("%d %d %lld", &N, &M, &K);
    long long ans = LLONG_MAX;
    for(int i=0;i<M;i++){
        long long u,v,w;
        scanf("%lld %lld %lld", &u, &v, &w);
        ans = min(ans, w);
    }
    printf("%lld\n", ans);
    return 0;
}

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