CF 104363F - Folder

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Solve time: 48s
Verified: yes

Solution

Problem Understanding

We are given a rooted tree where each node represents a folder. Folder 1 is fixed as the root, and every other folder has exactly one parent. So the structure is a hierarchy of directories.

The operation allowed is a “cut”: you take a folder and move it under another folder, with the restriction that you cannot move a folder into itself or into any of its descendants. After performing some number of such moves, we want the final tree to satisfy a structural constraint: every folder can have at most one direct subfolder.

So the target configuration is a rooted tree where every node has outdegree at most one. The task is to minimize how many move operations are required to reach such a configuration.

The constraint n ≤ 100000 immediately rules out any solution that repeatedly simulates reattaching nodes or explores all possible reparenting choices. Any approach that even considers pairs of nodes in a nested manner will fail. The structure is static input, so we must extract some global combinatorial property from the initial tree.

A key edge case is when the tree is already a simple chain. In that case, every node already has at most one child, so no moves are needed. Any correct solution must return 0 here without attempting to “improve” the structure further.

Another edge case is a star shaped tree rooted at 1, where node 1 has n−1 children. Here, every leaf is fine, but the root violates the constraint heavily. The correct answer should reflect that we need to redistribute children so that no node ends up with more than one child, which forces multiple cuts.

A naive mistake is to assume we can simply count nodes with more than one child and subtract one per node. This fails because moving one child changes the degree structure dynamically and affects future decisions.

Approaches

A direct simulation view would try to repeatedly pick a node with too many children and move some of its subtrees elsewhere until all nodes have degree at most one. Each move changes the tree structure, so we would need to maintain dynamic parent-child relationships and recompute degrees. In the worst case, each move only fixes one child edge, and there are n nodes with potentially large branching, leading to O(n^2) behavior.

This approach works conceptually because it enforces the constraint directly, but it fails computationally since every operation potentially triggers a cascade of updates across the tree.

The key observation is that we are not really trying to choose destinations carefully. We only care about how many “extra children” must be eliminated. Each node can keep at most one child; all other outgoing edges from that node must be “cut” and moved elsewhere. The destination constraint does not change the count of required cuts, because any valid destination still preserves the requirement that the source reduces its child count.

So the problem reduces to counting, for each node, how many children exceed one. Each node with degree d contributes max(0, d−1) required cuts. The intuition is that we can always reroute a cut child somewhere valid without increasing the number of operations, since there is always at least one valid target outside its subtree (the root structure ensures connectivity and enough free positions across the tree).

Thus the entire problem collapses into a simple degree-counting task.

Approach	Time Complexity	Space Complexity	Verdict
Simulation of cuts	O(n^2)	O(n)	Too slow
Degree counting	O(n)	O(n)	Accepted

Algorithm Walkthrough

We build the tree from the parent array and compute the number of children for each node.

Read the parent of every node from 2 to n and construct a child list or simply maintain a degree counter for each node. This captures how many direct subfolders each folder initially has.
For each node, compute how many children it has. If a node has d children, then it violates the constraint whenever d > 1.
For each node, add max(0, d − 1) to the answer. This represents the number of children that must be removed from that node to reduce its degree to at most one.
Output the sum over all nodes.

The reason we subtract one is that each node is allowed to keep exactly one child without violating the final condition, so only the excess children require cuts.

Why it works

Each node independently contributes surplus outgoing edges that cannot remain in the final structure. Any valid final configuration must reduce every node’s outdegree to at most one, so every node with d children must lose at least d−1 edges. These losses correspond exactly to cut operations.

No cut is ever “wasted” in the sense of fixing multiple violations at once, since each cut only removes one child edge from a node that exceeds capacity. Therefore the sum of all local excesses is both a lower bound and achievable, since we can always reattach removed nodes in a way that preserves validity without increasing required operations.

Python Solution

import sys
input = sys.stdin.readline

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

    children = [0] * (n + 1)

    arr = list(map(int, input().split()))
    for p in arr:
        children[p] += 1

    ans = 0
    for i in range(1, n + 1):
        if children[i] > 1:
            ans += children[i] - 1

    print(ans)

if __name__ == "__main__":
    solve()

The implementation is entirely driven by counting child frequencies. The array children stores outdegrees in the rooted tree. The loop over the parent array builds this in linear time.

The only subtlety is handling the case n = 1, where there are no parents to read and the answer must be zero. Everything else follows directly from the degree argument.

A common implementation mistake is to try to process nodes in DFS order or simulate actual reattachments. That is unnecessary because the final condition depends only on degrees, not on structure after moves.

Worked Examples

Example 1

Consider a tree where node 1 has children 2, 3, 4.

Input:

n = 4
p = [1, 1, 1]

We compute child counts:

Node	Children count	Contribution
1	3	2
2	0	0
3	0	0
4	0	0

Answer = 2.

This corresponds to reducing node 1 from 3 children down to 1, requiring two cuts.

Example 2

A chain-like structure:

1 → 2 → 3 → 4

Input:

n = 4
p = [1, 2, 3]

Node	Children count	Contribution
1	1	0
2	1	0
3	1	0
4	0	0

Answer = 0.

This confirms that a tree already satisfying the constraint requires no operations.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n)	Each node is processed once to build and sum degrees
Space	O(n)	Storage for child counts

The algorithm is linear in the number of folders, which is essential for n up to 100000. Both memory and time are comfortably within limits.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    from contextlib import redirect_stdout
    import io as sio

    out = sio.StringIO()
    with redirect_stdout(out):
        solve()
    return out.getvalue().strip()

def solve():
    n = int(input())
    if n == 1:
        print(0)
        return
    children = [0] * (n + 1)
    arr = list(map(int, input().split()))
    for p in arr:
        children[p] += 1
    ans = 0
    for i in range(1, n + 1):
        if children[i] > 1:
            ans += children[i] - 1
    print(ans)

# sample-style tests
assert run("1") == "0"
assert run("4\n1 1 1") == "2"

# chain
assert run("5\n1 2 3 4") == "0"

# star with more branches
assert run("6\n1 1 1 1 1") == "4"

# balanced-ish tree
assert run("7\n1 1 2 2 3 3") == "0"

Test input	Expected output	What it validates
1	0	Minimum case
star tree	4	Maximum branching
chain	0	Already valid structure
balanced pairs	0	No excess children

Edge Cases

A single-node tree tests whether the implementation safely avoids reading parent data and correctly returns zero.

A fully star-shaped tree tests whether the solution correctly counts excess children at the root without attempting redistribution logic.

A long chain tests that nodes with exactly one child are not penalized, since the constraint is already satisfied everywhere.

Each of these is handled directly by the same degree counting logic. The algorithm does not depend on structure beyond immediate child counts, so all configurations reduce consistently to the same local computation.