Overview

At first glance, this problem seems highly complex and solvable using Tree DP (Dynamic Programming on Trees). However, through mathematical observation (evaluating the boundary values), we can reduce it to an extremely simple problem: “just finding the minimum weight among all edges.”

---

Observation

1. Reformulating the condition “using at least two colors”

Coloring all sections with the same color is not allowed. This is equivalent to saying that there exists at least one edge in the tree whose endpoints have different colors.

2. Lower bound of the discomfort cost

Let \(e = (u, v)\) be an edge whose endpoints have different colors. If the colors assigned to the endpoints of this edge are \(a\) and \(b\) (\(a \neq b\)), the discomfort cost incurred at this edge is \(W_e \times |a - b|\).

Since \(a\) and \(b\) are distinct integers, the absolute difference between them is at least \(1\) (\(|a - b| \ge 1\)). Therefore, the cost incurred at this edge is at least \(W_e\).

Since the cost incurred at all other edges is at least \(0\), for any valid coloring (using at least two colors), the total discomfort cost is at least \(\min_{e} W_e\).

3. Achievability of the lower bound

Now, is it possible to make the total discomfort cost exactly \(\min_{e} W_e\)?

Let \(e_{\min}\) be the edge with the minimum weight. If we remove \(e_{\min}\) from the tree, the tree is divided into two connected components (subtrees). Let us color all sections in one connected component with color \(1\), and all sections in the other connected component with color \(2\).

In this case, the cost incurred at each edge is as follows: - Edge \(e_{\min}\): Since the colors of its endpoints are \(1\) and \(2\), the cost is \(W_{\min} \times |1 - 2| = W_{\min}\). - All other edges: Since both endpoints have the same color (both \(1\) or both \(2\)), the cost is \(0\).

In this coloring, the number of colors used is \(2\) (colors \(1\) and \(2\)), which is valid since \(K \ge 2\). Thus, the total discomfort cost in this case is exactly \(W_{\min}\).

4. Conclusion

From the above observations, it is shown that the minimum total discomfort cost among all valid colorings is indeed the minimum weight \(W_i\) of all given edges.

---

Algorithm

The provided code naively simulates and calculates this minimum value using Tree DP (Dynamic Programming).

DP Definition

• \(dp[u][c]\) : The minimum cost in the subtree rooted at \(u\), given that vertex \(u\) is colored with color \(c\).

Transitions

Consider the transition from a child node \(v\) to its parent node \(u\). If we compute the transition naively, we need to find \(g[c] = \min_{c'} (dp[v][c'] + W \times |c - c'|)\) for each color \(c\), which takes \(O(K^2)\) time with a standard loop.

In the provided code, to eliminate the absolute value symbol, the cumulative minimums from the left and right (arrays L and R) are precomputed, accelerating this transition to \(O(K)\): - \(L[c] = \min_{c' \le c} (dp[v][c'] - W \times c')\) - \(R[c] = \min_{c' \ge c} (dp[v][c'] + W \times c')\)

Using these, the transitioned cost \(g[c]\) for each \(c\) can be found in \(O(1)\) as follows: $\(g[c] = \min(W \times c + L[c], -W \times c + R[c])\)$

Then, the DP table update formula is as follows: dp[u][c] = min({ dp[u][c], g[c], W })

Here, the key point is taking the minimum with the edge weight \(W\) itself. This corresponds to the option of “switching colors using this edge as a boundary (paying the cost \(W\))”. By performing this operation bottom-up from the leaves, the value of dp[0][c] at the root eventually converges to the minimum edge weight of the entire tree, \(\min W_i\).

---

Complexity

Complexity of the provided DP code

• Time Complexity: \(O(N K)\) Because we run a loop over the number of colors \(K\) at each vertex.
• Space Complexity: \(O(N K)\) To store the DP table dp[N][K].

Note: Under the constraint \(K \le 10^8\), the provided code will exceed the memory limit (MLE) and time limit (TLE). However, by utilizing the observation above (simply outputting the minimum value), we can dramatically speed it up as follows.

Complexity of the optimized approach

Since we only need to find the minimum weight among all edges, there is no need to even construct the graph. - Time Complexity: \(O(N)\) (only for reading the input) - Space Complexity: \(O(1)\)

---

Implementation Points

1. Transition optimization technique (Cumulative \(\min\) from left and right)

The transition using L and R in the provided code is a typical optimization technique in “Tree DP with constant-slope cost functions”. It is a very powerful method when \(K\) is small.

2. Ultimate Simplification

To satisfy the constraint \(K \le 10^8\) and get an Accepted (AC) verdict, instead of implementing DP, it is sufficient to directly output the minimum weight of all edges as shown below.

#include <iostream>
#include <algorithm>

using namespace std;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int N, M;
    long long K;
    if (!(cin >> N >> M >> K)) return 0;

    long long min_W = 1e18; // Initialize with a sufficiently large value
    for (int i = 0; i < M; ++i) {
        int u, v;
        long long w;
        cin >> u >> v >> w;
        min_W = min(min_W, w);
    }

    cout << min_W << "\n";
    return 0;
}

Source Code

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

const long long INF = 1e18;

int N, M, K;
vector<vector<pair<int, long long>>> adj;
vector<vector<long long>> dp;

void dfs(int u, int p) {
    for (auto [v, W] : adj[u]) {
        if (v == p) continue;
        dfs(v, u);
        
        vector<long long> L(K), R(K), g(K);
        L[0] = dp[v][0];
        for (int c = 1; c < K; ++c) {
            L[c] = min(L[c-1], dp[v][c] - W * c);
        }
        R[K-1] = dp[v][K-1] + W * (K-1);
        for (int c = K-2; c >= 0; --c) {
            R[c] = min(R[c+1], dp[v][c] + W * c);
        }
        for (int c = 0; c < K; ++c) {
            g[c] = min(W * c + L[c], -W * c + R[c]);
        }
        
        for (int c = 0; c < K; ++c) {
            dp[u][c] = min({ dp[u][c], g[c], W });
        }
    }
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    
    if (!(cin >> N >> M >> K)) return 0;
    
    adj.resize(N);
    for (int i = 0; i < M; ++i) {
        int u, v;
        long long w;
        cin >> u >> v >> w;
        --u; --v;
        adj[u].push_back({v, w});
        adj[v].push_back({u, w});
    }
    
    dp.assign(N, vector<long long>(K, INF));
    
    dfs(0, -1);
    
    long long ans = INF;
    for (int c = 0; c < K; ++c) {
        ans = min(ans, dp[0][c]);
    }
    
    cout << ans << "\n";
    
    return 0;
}

---

This editorial was generated by gemini-3.5-flash-high.