Overview

This is an optimization problem where we decide, for each edge in a rooted tree, which of the two settlements at its endpoints should be responsible for it. Each settlement has a capacity, and exceeding it incurs additional costs. We find the minimum cost by combining tree DP with a greedy approach.

Analysis

Key Observations

1. Independent choice per edge: For each edge \((i, P_i)\), we must decide between “assign to parent \(P_i\)” or “assign to child \(i\)” — a binary choice for every edge.

2. Tree DP formulation: For each settlement, we need to determine “how many edges it is responsible for,” and we can build the optimal solution on a subtree-by-subtree basis.

3. Marginal cost structure: When incrementally assigning one more edge to settlement \(i\), the penalty is 0 up to the \(C_i\)-th edge, and \(W_i\) per edge from the \((C_i + 1)\)-th edge onward. Since this marginal cost is monotonically non-decreasing, a greedy approach is applicable.

DP State Definition

Define \(f[v][b]\) as “the minimum additional cost (penalty) incurred within the subtree of settlement \(v\)”: - \(b = 0\): Edge \((v, P_v)\) is handled by parent \(P_v\) - \(b = 1\): Edge \((v, P_v)\) is handled by settlement \(v\)

Algorithm

Basic Approach

For each child \(c\) of node \(v\), edge \((v, c)\) can be: - Assigned to child \(c\) → subtree cost from child is \(f[c][1]\) - Assigned to parent \(v\) → subtree cost from child is \(f[c][0]\), and \(v\)’s responsibility count increases by 1

Greedy Switching

1. Initial state: Start from the state where all child edges are assigned to the child side (cost \(= \sum_c f[c][1]\))
2. Switching gain: Moving child \(c\)’s edge to \(v\) reduces the subtree cost by \(d = f[c][1] - f[c][0]\)
3. Marginal penalty: If \(v\)’s current responsibility count is \(\geq C_v\), the penalty is \(W_v\); otherwise it is \(0\)
4. Decision: Continue switching as long as gain \(d\) exceeds marginal penalty \(mp\)

Process children in decreasing order of \(d\), and stop when \(d \leq mp\) (since \(d\) is sorted in descending order and \(mp\) is non-decreasing, this is optimal).

Formula

\[f[v][b] = \left(\sum_c f[c][1]\right) - \sum_{\text{switched children}} (d - mp) + W_v \cdot \max(0, b - C_v)\]

The final answer is \((N-1) + f[0][0]\).

Complexity

• Time complexity: \(O(N \log N)\) (total cost of sorting the children lists at each node)
• Space complexity: \(O(N)\)

Implementation Notes

• Special handling of the root: The root (settlement 1) has no parent edge, so the state \(b = 1\) does not exist. In the code, we set f[0][1] = 0 and continue.

• BFS-order processing: Instead of recursion, we use a stack to determine topological order and process in reverse order (from leaves to root) to avoid stack overflow.

• Greedy via sorting: By sorting \(d = f[c][1] - f[c][0]\) in descending order, we prioritize assigning edges with the largest gain first, and the correctness of stopping when the marginal cost exceeds the gain is guaranteed.

  Source Code

import sys
from sys import setrecursionlimit

def main():
    input_data = sys.stdin.buffer.read().split()
    idx = 0
    N = int(input_data[idx]); idx += 1
    children = [[] for _ in range(N)]
    for i in range(1, N):
        p = int(input_data[idx]) - 1; idx += 1
        children[p].append(i)
    C = [0] * N; W = [0] * N
    for i in range(N):
        C[i] = int(input_data[idx]); idx += 1
        W[i] = int(input_data[idx]); idx += 1

    f = [[0, 0] for _ in range(N)]
    order = []
    stack = [0]
    while stack:
        v = stack.pop()
        order.append(v)
        for c in children[v]:
            stack.append(c)
    for v in reversed(order):
        base = sum(f[c][1] for c in children[v])
        ds = sorted([f[c][1] - f[c][0] for c in children[v]], reverse=True)
        for b in range(2):
            if v == 0 and b == 1:
                f[v][1] = 0
                continue
            cost = base
            taken = b
            benefit = 0
            for d in ds:
                marginal_penalty = W[v] if taken >= C[v] else 0
                net = d - marginal_penalty
                if net > 0:
                    benefit += net
                    taken += 1
                else:
                    break
            cost -= benefit
            cost += W[v] * max(0, b - C[v]) if b > 0 else 0
            # Recalculate properly
            total = base
            taken = b
            for d in ds:
                mp = W[v] if taken >= C[v] else 0
                if d > mp:
                    total -= d; total += mp
                    taken += 1
                else:
                    break
            f[v][b] = total + W[v] * max(0, b - C[v])
    print((N - 1) + f[0][0])

main()

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