CF 1714G - Path Prefixes

Rating: 1700
Tags: binary search, data structures, dfs and similar, trees
Solve time: 2m 11s
Verified: no

Solution

Problem Understanding

We are asked to process a rooted tree with values on each edge. Each edge has two integers, $a_j$ and $b_j$. For every non-root node $i$, we consider the path from the root to $i$. Let $A_i$ be the sum of all $a_j$ along this path. We are asked to find $r_i$, the length of the longest prefix of this path such that the sum of $b_j$ along the prefix does not exceed $A_i$. The prefix length is measured in edges, so $r_i = 0$ means no edge can be included in the prefix.

The input gives multiple test cases, each describing a tree with $n$ nodes and $n-1$ edges, where each edge is defined by a parent $p_j$ and the two costs $a_j$ and $b_j$. The output is an array of $n-1$ integers, one per non-root node.

The constraints are tight: $n$ can be up to $2 \cdot 10^5$, and the sum of $n$ over all test cases is also bounded by $2 \cdot 10^5$. A naive approach that explicitly sums all prefixes for each node would be $O(n^2)$, which is far too slow. We need a solution linear or near-linear in the number of nodes per test case. Edge cases arise when all $b_j$ are larger than $A_i$ for a node, in which case the correct prefix length is zero, and when $A_i$ exactly matches some prefix sum, which must be handled carefully to avoid off-by-one errors.

Approaches

A brute-force approach would be to reconstruct the path from the root to each node $i$ and then iterate over all prefixes to compute the sum of $b_j$, stopping when the sum exceeds $A_i$. This is correct but requires $O(n^2)$ operations in the worst case, which fails the largest constraints.

The key observation for optimization is that we only need the prefix sums of $b_j$ along the path from the root. If we maintain an array that stores the cumulative sum of $b_j$ values along the path to the current node, we can binary search over these sums to find the longest prefix whose sum does not exceed $A_i$. Because the tree is rooted, we can build these prefix sums using a depth-first search, propagating cumulative sums from parent to child. Each binary search operation takes $O(\log n)$, and visiting each node takes $O(1)$, leading to an overall $O(n \log n)$ solution. Using a list to store cumulative sums in DFS ensures we can access ancestors efficiently.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(n^2)	O(n)	Too slow
DFS + Binary Search on prefix sums	O(n log n)	O(n)	Accepted

Algorithm Walkthrough

Initialize an array prefix_b to store cumulative sums of $b_j$ along the path from the root. The root node has a cumulative sum of zero. Maintain an array A to track cumulative sums of $a_j$ from the root.
Construct the tree as an adjacency list, storing for each child the edge values $a_j$ and $b_j$. This allows easy DFS traversal from the root.
Start DFS from the root. For each node $u$ visited, compute A[u] = A[parent] + a_edge, which is the total sum of $a_j$ along the path from the root to $u$.
Maintain a prefix_b array representing the cumulative sum of $b_j$ along the path. When visiting a child, append prefix_b[-1] + b_edge to the array before recursive DFS call.
To compute $r_i$ for the current node, perform a binary search over prefix_b to find the rightmost index idx such that prefix_b[idx] <= A[i]. This index is exactly the length of the longest prefix.
After finishing DFS for a child, pop the last element from prefix_b to backtrack and restore the state for the next sibling.

Why it works: At each node, the prefix_b array stores cumulative sums of $b_j$ along the path from the root. Because the tree is rooted, the path is unique, and the array is strictly increasing. Binary searching over this array ensures we find the longest prefix satisfying the sum constraint. The DFS traversal guarantees each node is visited exactly once and the prefix sums reflect the correct path.

Python Solution

import sys
input = sys.stdin.readline
import bisect

def solve():
    t = int(input())
    for _ in range(t):
        n = int(input())
        tree = [[] for _ in range(n + 1)]
        for j in range(2, n + 1):
            p, a, b = map(int, input().split())
            tree[p].append((j, a, b))

        A = [0] * (n + 1)
        res = [0] * (n + 1)
        prefix_b = [0]

        def dfs(u):
            for v, a_edge, b_edge in tree[u]:
                A[v] = A[u] + a_edge
                prefix_b.append(prefix_b[-1] + b_edge)
                # Binary search to find the longest prefix
                idx = bisect.bisect_right(prefix_b, A[v]) - 1
                res[v] = idx
                dfs(v)
                prefix_b.pop()

        dfs(1)
        print(" ".join(str(res[i]) for i in range(2, n + 1)))

if __name__ == "__main__":
    solve()

The DFS maintains the prefix_b array as the cumulative sum of $b_j$ along the path to the current node. Binary search is used to efficiently find the maximum prefix length for each node, and popping after visiting children restores the correct state for sibling traversal. Using bisect_right ensures we handle the case where the prefix sum equals A_i correctly.

Worked Examples

Sample 1

Node	Path to Node	A_i	prefix_b	r_i
2	1→2	5	[0,6]	0
3	1→2→3→4	19	[0,6,16,17]	3
4	1→2→3	14	[0,6,16]	1
5	1→2→3→5	16	[0,6,16,17]	2

The table confirms that the DFS with cumulative sums produces the correct longest prefix length for each node.

Sample 2

Node	Path to Node	A_i	prefix_b	r_i
2	1→2	1	[0,100]	0
3	1→2→3	2	[0,100,101]	0
4	1→2→3→4	103	[0,100,101,102]	3

This shows the edge case where the first edge $b_j > A_i$ is handled correctly, giving prefix length 0.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n log n)	DFS visits each node once, binary search per node takes log(depth), total O(n log n)
Space	O(n)	Tree adjacency list, cumulative sums array and result array

The algorithm fits well within the 2-second time limit and 256 MB memory constraint, since $n \le 2\cdot 10^5$ and each node contributes only log(n) operations.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    from contextlib import redirect_stdout
    out = io.StringIO()
    with redirect_stdout(out):
        solve()
    return out.getvalue().strip()

# provided samples
assert run("""4
9
1 5 6
4 5 1
2 9 10
4 2 1
1 2 1
2 3 3
6 4 3
8 1 3
4
1 1 100
2 1 1
3 101 1
4
1 100 1
2 1 1
3 1 101
10
1 1 4
2 3 5
2 5 1
3 4 3
3 1 5
5 3 5
5 2 1
1 3 2
6 2 1
""") == "0 3 1 2 1 1 2 3\n0 0 3\n1 2 2\n0 1 2 1 1 2 2 1 1",