Overview

This problem asks us to find the minimum sum of costs of all edges in a tree when we set the multiplier \(r_i\) for each edge to satisfy the given conditions. Although it looks like a complex optimization problem at first glance, clarifying the conditions reveals that all multipliers are uniquely determined to be \(1\). Therefore, the minimum value we are looking for is simply the sum of the weights \(w_i\) of all edges.

Analysis

Key Insight: Focus on paths of length \(1\)

Let us focus on the following condition in the problem description:

For any pair of distinct vertices in the tree, let \(k\) be the number of edges in the path between them. Then, the following must hold: \(r_{e_1} + r_{e_2} + \cdots + r_{e_k} \leq k^2\).

This condition must hold for any pair of distinct vertices. Therefore, let us consider the simplest case: a path connecting two adjacent vertices (a path of length \(k=1\)).

For any edge \(i\) in the tree, the path consisting of only that edge has length \(k=1\). Applying the condition to this path yields: $\(r_i \leq 1^2 = 1\)$

On the other hand, according to the problem statement, the multiplier \(r_i\) must be a positive integer (an integer greater than or equal to \(1\)). That is, $\(r_i \geq 1\)$ holds.

The only integer \(r_i\) that satisfies both inequalities simultaneously is \(r_i = 1\).

Consistency when setting all \(r_i = 1\)

When we set \(r_i = 1\) for all edges, the sum of multipliers along any path of length \(k\) is: $\(r_{e_1} + r_{e_2} + \cdots + r_{e_k} = \underbrace{1 + 1 + \cdots + 1}_{k \text{ elements}} = k\)$

For any positive integer \(k \geq 1\), \(k \leq k^2\) always holds. Therefore, if we set the multiplier of all edges to \(1\), we can confirm that all conditions for paths of length \(2\) or greater are also automatically satisfied.

Conclusion

There is only \(1\) valid assignment of multipliers that satisfies the conditions, which is \(r_i = 1 \ (1 \leq i \leq N-1)\). Therefore, the minimum sum of costs of all edges is simply the sum of the weights of all edges: $\(\sum_{i=1}^{N-1} w_i \times 1 = \sum_{i=1}^{N-1} w_i\)$

Algorithm

1. Read the number of vertices \(N\) and the information of each edge (\(u_i, v_i, w_i\)) from the input.
2. There is no need to construct the connections between vertices (such as an adjacency list). Just sum up the weight \(w_i\) of each edge.
3. Output the calculated sum.

Complexity

• Time Complexity: \(O(N)\) Since we only need to iterate through the weights of the \(N-1\) edges and sum them up, the process is completed in linear time \(O(N)\).
• Space Complexity: \(O(N)\) The memory required to store the input is \(O(N)\). It can also be reduced to \(O(1)\) if the input is processed on the fly as a stream.

Key Points for Implementation

• Handling Overflow: Since \(N\) is at most \(2 \times 10^5\) and the weight \(w_i\) is at most \(10^9\), the maximum possible value of the answer is around \(2 \times 10^{14}\). In Python, arbitrary-precision integers are used automatically, so you do not need to worry about it. However, when using languages like C++, make sure to use a 64-bit integer type (such as long long) for the variable storing the answer.

  Source Code

import sys

def solve():
    input_data = sys.stdin.read().split()
    if not input_data:
        return
    N = int(input_data[0])
    ans = 0
    for i in range(N - 1):
        ans += int(input_data[3 * i + 3])
    print(ans)

if __name__ == '__main__':
    solve()

---

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