Overview

This problem involves a tree-structured road network where, for each edge (road), we must decide which of the two adjacent settlements is responsible for the waterway construction, minimizing the total additional cost incurred. By leveraging the properties of the tree structure, we can efficiently solve this using Tree DP (Dynamic Programming), which determines optimal assignments bottom-up from the leaves toward the root.

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Analysis

1. Structuring the Decision

Each road (edge) connects a parent settlement and a child settlement. There are only two options for who takes charge of the road’s construction: “the parent takes responsibility” or “the child takes responsibility.” Therefore, for each subtree, we can classify states based on “which side is responsible for the road connecting the subtree’s root to its parent.”

Specifically, for each settlement \(u\), we define the following two states (minimum costs): - \(dp_0[u]\): The minimum additional cost within the subtree of settlement \(u\) when the road connecting settlement \(u\) and its parent \(P_u\) is handled by the parent \(P_u\). - \(dp_1[u]\): The minimum additional cost within the subtree of settlement \(u\) when the road connecting settlement \(u\) and its parent \(P_u\) is handled by the child \(u\).

2. Transitions from Child Settlements

Let \(Ch(u)\) denote the set of child settlements of settlement \(u\). Assume that settlement \(u\) takes responsibility for exactly \(c\) of the \(k\) roads connecting it to its children (\(0 \leq c \leq k\)).

For the \(c\) selected child settlements \(v\), since \(u\) handles the road \((u, v)\), from child \(v\)’s perspective “the parent handles it,” so the cost is \(dp_0[v]\). For the remaining \(k - c\) unselected child settlements \(v\), since \(v\) handles the road \((u, v)\), the cost is \(dp_1[v]\).

Therefore, if we let the set of child settlements handled by settlement \(u\) be \(S \subseteq Ch(u)\) (where \(|S| = c\)), the total cost from child settlements is: $\( \sum_{v \in S} dp_0[v] + \sum_{v \notin S} dp_1[v] \)$

This can be rewritten as: $\( \sum_{v \in Ch(u)} dp_1[v] + \sum_{v \in S} (dp_0[v] - dp_1[v]) \)$

3. Optimization via Greedy Selection

To minimize the above value for any given \(c\), we should select the \(c\) children with the smallest differences \(D[v] = dp_0[v] - dp_1[v]\) to include in \(S\). By sorting \(D[v]\) in ascending order beforehand, we can compute the minimum cost for all \(c \in [0, k]\) in \(O(k \log k)\) time.

The total number of roads \(m_u\) that settlement \(u\) itself handles is: - For \(dp_0[u]\) (parent handles the road to the parent): \(m_u = c\) - For \(dp_1[u]\) (\(u\) handles the road to the parent): \(m_u = c + 1\)

For each case, we add the additional cost incurred by settlement \(u\) itself, \(W_u \times \max(0, m_u - C_u)\), and take the minimum over all values of \(c\) to update \(dp_0[u]\) and \(dp_1[u]\).

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Algorithm

1. Graph Construction: Build the children list for each vertex from the given parent relationships.
2. DP Execution:
   • Process from the leaves (vertices with no children) toward the root. Since the constraint guarantees \(P_i < i\), iterating vertex numbers from \(N\) down to \(1\) allows bottom-up (topological order) processing.
   • For each vertex \(u\):
     1. For each child settlement \(v\), compute the difference \(D[v] = dp_0[v] - dp_1[v]\) and sort them in ascending order.
     2. Loop from \(c = 0\) to \(c = k\) (total number of children), using prefix sums to find the “minimum cost when selecting \(c\) children.”
     3. For each case of \(dp_0[u]\) and \(dp_1[u]\), add the additional cost corresponding to the number of roads \(m_u\) that the settlement handles, and record the minimum value.
3. Output the Answer: Since the root (vertex 1) has no parent, \(dp_0[1]\)—which treats the road above the root as “handled by the parent (i.e., non-existent)“—gives the minimum additional cost. Output this value plus the base construction cost \(N - 1\).

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Complexity

Time Complexity: \(O(N \log N)\)

For each vertex \(u\) with \(k_u\) children, sorting takes \(O(k_u \log k_u)\) time. The sum of \(k_u\) over all vertices equals \(N - 1\), the number of edges in the tree. Therefore, the total time for sorting is \(\sum_{u=1}^{N} O(k_u \log k_u) \leq O(N \log N)\), which is sufficiently fast.

Space Complexity: \(O(N)\)

The memory required for the adjacency list representing the tree structure and the DP tables (dp0, dp1) is \(O(N)\).

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Implementation Notes

• Bottom-up processing without recursion: In languages such as Python, performing depth-first search (DFS) with recursive functions may hit the maximum recursion depth limit or incur significant overhead. In this problem, since the parent’s number is always smaller than the child’s (\(P_i < i\)), simply iterating from \(N\) down to \(1\) allows us to implement bottom-up tree DP safely and efficiently without recursion.

• Adding the base construction cost: Be careful not to forget adding the “base construction cost \(1 \times (N-1)\)” mentioned in the problem statement at the end.

  Source Code

import sys


def solve():
    input = sys.stdin.read
    data = input().split()
    if not data:
        return

    N = int(data[0])
    P = [0, 0] + [int(x) for x in data[1:N]]

    C = [0] * (N + 1)
    W = [0] * (N + 1)
    idx = N
    for i in range(1, N + 1):
        C[i] = int(data[idx])
        W[i] = int(data[idx + 1])
        idx += 2

    children = [[] for _ in range(N + 1)]
    for i in range(2, N + 1):
        children[P[i]].append(i)

    dp0 = [0] * (N + 1)
    dp1 = [0] * (N + 1)

    for u in range(N, 0, -1):
        ch = children[u]
        k = len(ch)
        if k == 0:
            dp0[u] = 0
            diff = 1 - C[u]
            dp1[u] = W[u] * (diff if diff > 0 else 0)
            continue

        D = [dp0[v] - dp1[v] for v in ch]
        D.sort()

        sum_dp1 = 0
        for v in ch:
            sum_dp1 += dp1[v]

        curr_S = sum_dp1
        diff_C0 = -C[u]
        diff_C1 = 1 - C[u]
        min_val0 = curr_S + W[u] * (diff_C0 if diff_C0 > 0 else 0)
        min_val1 = curr_S + W[u] * (diff_C1 if diff_C1 > 0 else 0)

        for c in range(1, k + 1):
            curr_S += D[c - 1]
            diff0 = c - C[u]
            diff1 = c + 1 - C[u]
            val0 = curr_S + W[u] * (diff0 if diff0 > 0 else 0)
            val1 = curr_S + W[u] * (diff1 if diff1 > 0 else 0)
            if val0 < min_val0:
                min_val0 = val0
            if val1 < min_val1:
                min_val1 = val1

        dp0[u] = min_val0
        dp1[u] = min_val1

    ans = (N - 1) + dp0[1]
    print(ans)


if __name__ == "__main__":
    solve()

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