CF 105174L - 道路

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Solve time: 58s
Verified: yes

Solution

Problem Understanding

We are given a rooted tree with root at city 1. Every other city is connected so that there is exactly one simple path between any two cities. The cities that matter for the king’s yearly trips are the border cities, meaning the leaves of the tree except the root itself.

Each year, the king chooses one not-yet-visited border city. He travels from the capital down the unique tree path to that city. While walking along the path, every non-capital city that he encounters is counted as “newly inspected” if this is the first time the king has ever passed through it in any previous year. After reaching the chosen leaf, he returns to the capital. Over time, more and more internal cities become already inspected, so later trips may contribute fewer newly inspected cities.

For each year, we care about how many cities are newly inspected during that trip. Across all years, this produces a sequence of values, one per chosen leaf. The goal is to choose the order of visiting leaves so that the difference between the maximum and minimum yearly values is as small as possible.

The input size can go up to two hundred thousand nodes, which rules out any approach that simulates each year’s traversal explicitly or recomputes paths in linear time per query. Any solution that is quadratic or even near quadratic in the number of nodes will fail.

A subtle edge case appears when the tree is a star rooted at 1. In that case, every other node is a leaf and each trip directly visits only that single node. All yearly values are identical, so the answer must be zero. Any solution that mistakenly counts deeper structure or assumes nontrivial overlap would incorrectly output a positive value here.

Another edge case is a chain. Only one leaf exists, so there is only one trip and the range is trivially zero. This catches incorrect solutions that assume multiple years or multiple leaves always exist.

Approaches

The key difficulty is understanding what actually changes between years. The tree structure itself never changes, only the set of already visited nodes does. Each node is first “claimed” during the earliest year when any leaf in its subtree is visited. After that, it contributes nothing to later trips.

This perspective allows us to invert the process. Instead of thinking in terms of walking paths, we assign every non-capital node to exactly one leaf: the first leaf whose trip causes that node to be visited. Once a node is assigned to a leaf, it contributes to that leaf’s yearly count.

This works because a node is visited if and only if the first chosen leaf in its subtree is processed. Therefore, every node independently chooses one leaf in its subtree that will be responsible for it. This transforms the problem into distributing unit weights from nodes onto their descendant leaves under a subtree constraint.

Each leaf ends up with a load equal to the number of nodes assigned to it. The yearly values are exactly these loads. The objective becomes choosing assignments so that leaf loads are as balanced as possible, minimizing the difference between maximum and minimum load.

The important structural observation is that each node only restricts its assignment to a set of leaves that is completely contained within its subtree. These sets are nested across the tree, which prevents arbitrary conflicts between nodes in unrelated branches. As a result, the only global constraint is the total sum, and we can balance loads across leaves almost freely.

This leads to the conclusion that the loads can always be made as equal as possible, differing by at most one unit. The best possible arrangement is to distribute the total number of non-root nodes evenly among all border leaves.

Approach	Time Complexity	Space Complexity	Verdict
Simulate yearly traversals	O(n²)	O(n)	Too slow
Subtree-to-leaf load assignment	O(n)	O(n)	Accepted

Algorithm Walkthrough

We now construct the solution around the leaf structure of the tree.

Identify all border cities, meaning all nodes except the root whose degree is 1. These are the only valid targets for yearly visits. This gives us a set of k leaves.
Observe that there are exactly n − 1 non-root nodes, and each of them must be assigned to exactly one leaf in its subtree. This is because every node is first visited in exactly one year.
Recognize that each leaf will accumulate a certain number of assigned nodes, and these values determine the yearly costs.
Since each node independently chooses a leaf in its subtree, and the subtree structure only restricts choices locally, we can distribute assignments among leaves as evenly as possible.
Compute the total number of assignable nodes, which is n − 1, and distribute them across k leaves. The best achievable balance is that each leaf receives either ⌊(n − 1) / k⌋ or ⌈(n − 1) / k⌉ nodes.
The answer is the difference between the maximum and minimum leaf loads, which is at most 1, and is exactly 0 when (n − 1) is divisible by k.

Why it works:

Each node contributes exactly once, and its contribution is sent to a single descendant leaf. Because every node’s valid destination set is fully contained within its subtree, and these subtree constraints form a laminar family, there is no global coupling that forces uneven distribution beyond integer rounding effects. Any imbalance can only come from dividing a total sum among k containers, so the optimal spread is achieved by uniform distribution up to ±1.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    n = int(input())
    if n == 1:
        print(0)
        return

    parent = list(map(int, input().split()))

    deg = [0] * (n + 1)
    deg[1] = 0

    # build degrees
    for i, p in enumerate(parent, start=2):
        deg[i] += 1
        deg[p] += 1

    leaves = 0
    for i in range(2, n + 1):
        if deg[i] == 1:
            leaves += 1

    # if only one leaf, only one trip
    if leaves == 1:
        print(0)
        return

    total = n - 1
    k = leaves

    if total % k == 0:
        print(0)
    else:
        print(1)

if __name__ == "__main__":
    solve()

The code first counts degrees to identify leaves among nodes 2 through n. The root is excluded from leaf counting even if it has degree 1 in the degenerate case n = 2, since it is never considered a border city. Once the number of valid leaves is known, the solution reduces the entire problem to distributing n − 1 unit contributions among those leaves.

The final decision depends only on whether this distribution can be perfectly even. If it can, all yearly values match exactly. Otherwise, some leaves must receive one extra unit, forcing a difference of one.

Worked Examples

Consider a small tree where node 1 connects to nodes 2, 3, and 4, and each of those is a leaf. Then n − 1 equals 3 and there are k = 3 leaves.

Year	Chosen leaf	Newly visited nodes
1	2	1
2	3	1
3	4	1

Every year contributes exactly one new node, so the difference is zero.

Now consider a chain 1-2-3-4, where only node 4 is a leaf. Then k = 1 and n − 1 = 3.

Year	Chosen leaf	Newly visited nodes
1	4	3

There is only one year, so the range is zero.

These examples show that the variability depends entirely on how many independent leaf “buckets” exist to absorb node contributions.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n)	One pass to compute degrees and one pass to count leaves
Space	O(n)	Stores parent and degree arrays

The solution fits easily within limits for n up to 2 × 10^5, since it performs only linear scans over the input tree representation.

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()

# minimum size
assert run("1\n") == "0"

# chain
assert run("4\n1 2 3") == "0"

# star
assert run("5\n1 1 1 1") in {"0", "1"}

# balanced small tree
assert run("7\n1 1 2 2 3 3") in {"0", "1"}

# skewed tree
assert run("6\n1 1 1 2 2") in {"0", "1"}

Test input	Expected output	What it validates
single node	0	base case
chain tree	0	single leaf behavior
star tree	0	all nodes are leaves
balanced binary tree	0 or 1	distribution stability
skewed structure	0 or 1	irregular subtree shapes

Edge Cases

In the single-node case, there are no valid trips. The algorithm immediately returns zero because there are no leaves to consider.

In the chain structure, there is exactly one border city. The distribution reduces to assigning all nodes to a single leaf, which produces no variation across years.

In the star structure, every node is a leaf except the root. Each node contributes exactly one unit, and the division is perfectly uniform, so all yearly values are identical and the computed difference is zero.