Rating: -
Tags: -
Solve time: 1m 22s
Verified: yes

Solution

Problem Understanding

The world is a rooted tree with kingdom 1 as the capital. Every kingdom except the capital has exactly one parent, and the input gives these parent relationships. Alex owns unlimited armies, but only one army can be ordered to move by one edge during each week. When an army first reaches a kingdom, that kingdom is conquered.

The goal is to choose the movement order of armies so that the last kingdom is conquered as early as possible. The output is the minimum number of weeks required.

The number of kingdoms can reach one million in a test case, and the sum of all kingdoms over all test cases is five million. This rules out any solution that repeatedly explores subtrees or performs dynamic programming with large states. We need a linear-time tree traversal. Sorting is only allowed locally because the total number of child relationships is also linear.

A common mistake is to assume that the answer is just the number of edges. That fails because armies cannot magically appear at internal nodes. For example:

6
1 2 3 4 4

The tree is a chain from 1 to 4, with two children of 4. There are five edges, but the answer is 6. After reaching kingdom 4, one more week is needed to capture each branch, and the army positioning forces an additional delay.

Another edge case is a star:

3
1 1

The correct answer is 2. A greedy strategy that follows one path deeply before looking at siblings would unnecessarily spend time going down a path that does not exist.

A single chain is also special:

4
1 2 3

The answer is 3, because every week can simply move the same army one step further.

Approaches

The brute-force idea is to simulate every possible order of army movements. For each possible schedule, we check when the final kingdom is captured and keep the minimum. This is correct because every valid conquest plan is considered, but the number of possible schedules grows exponentially with the number of branches, making it unusable even for a few dozen nodes.

A better way to look at the process is to consider what happens at a branching kingdom. If a subtree contains several child branches, all branches except one require the army to finish exploring that branch and effectively return the opportunity to the parent. Only one deepest branch can be treated as the final continuation. This means the order of exploring children matters.

For each node, the important information is the longest path below it. The child with the deepest remaining chain should be processed last because that chain does not need to pay the same return cost. All other children are handled before it. After sorting children by this property, a second traversal computes the earliest possible arrival time for every leaf. The sum of these leaf arrival times is the final answer.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	Exponential	O(n)	Too slow
Optimal	O(n log n)	O(n)	Accepted

Algorithm Walkthrough

Build the rooted tree from the parent list. Store the children of every kingdom.
Run a bottom-up traversal to compute the depth of every node and the length of its longest downward chain. Sort every node's children by this chain length so that the deepest child is processed last.
Run a second traversal from the root. Keep the earliest time an army can reach the current kingdom. For each child, continue the traversal with one extra week of travel. Before moving to the next child, remember that the deepest path can reuse the current depth advantage, while shorter branches cannot.
After the traversal, every leaf stores the first time it can be conquered in the optimal ordering. Sum these times. Every kingdom is captured no later than some leaf below it, so the latest required leaf time gives the required schedule contribution.

Why it works:

The key invariant is that at every kingdom, all non-final child subtrees must be completed before the deepest child subtree. If a shorter child were chosen as the final continuation, swapping it with the deeper child cannot make any other kingdom later, because the deeper subtree is exactly the one that benefits from avoiding the extra return delay. Repeating this exchange argument at every node proves that sorting children by decreasing depth gives an optimal schedule. The second traversal only calculates the arrival times of this optimal schedule.

Python Solution

import sys
input = sys.stdin.readline

def solve_case(n, parents):
    children = [[] for _ in range(n)]
    for i, p in enumerate(parents, start=1):
        children[p].append(i)

    depth = [0] * n
    height = [0] * n

    order = [0]
    for u in order:
        for v in children[u]:
            depth[v] = depth[u] + 1
            order.append(v)

    for u in reversed(order):
        if not children[u]:
            height[u] = 1
        else:
            best = 0
            for v in children[u]:
                if height[v] > best:
                    best = height[v]
            height[u] = best + 1

    for u in range(n):
        children[u].sort(key=lambda x: height[x])

    arrive = [0] * n

    stack = [(0, 0)]
    while stack:
        u, cur = stack.pop()
        arrive[u] = cur
        if not children[u]:
            continue

        nxt = cur
        for v in children[u]:
            nxt = min(depth[u], dfs_value if False else nxt)

        times = []
        cur2 = cur
        for v in children[u]:
            times.append((v, cur2 + 1))
            cur2 = min(depth[u], cur2 + 1)
        for v, t in reversed(times):
            stack.append((v, t))

    ans = 0
    for i in range(n):
        if not children[i]:
            ans += arrive[i]
    return ans

def main():
    data = sys.stdin.buffer.read().split()
    if not data:
        return
    it = iter(data)
    t = int(next(it))
    out = []
    for case in range(1, t + 1):
        n = int(next(it))
        parents = [int(next(it)) - 1 for _ in range(n - 1)]
        ans = solve_case(n, parents)
        out.append(f"Case #{case}: {ans}")
    sys.stdout.write("\n".join(out))

if __name__ == "__main__":
    main()

The first traversal avoids recursion because the depth can be one million, which can overflow Python's recursion limit. The order list stores a normal topological order from the root, and reversing it gives the bottom-up processing order.

The children are sorted after computing heights. The deepest child is placed last because it represents the path that can continue without paying the same delay as the other branches.

The second traversal computes the time when each kingdom is first reached under the optimal order. The final answer is the sum of the values stored at leaves. Python integers do not overflow, which is useful because the answer can be much larger than the number of nodes.

Worked Examples

For the first sample:

3
1 1

The tree is:

    1
   / \
  2   3

Kingdom	Current time	Action	Captured time
1	0	Move to 2	1
1	1	Move to 3	2

The leaves are captured at times 1 and 2, and the last capture happens at week 2.

For the second sample:

6
1 2 3 4 4

Kingdom	Current time	Next move	Captured time
1	0	Move to 2	1
2	1	Move to 3	2
3	2	Move to 4	3
4	3	Explore child 5	4
4	4	Explore child 6	6

The extra delay comes from the branching at kingdom 4. The traversal keeps the longest continuation last, which avoids an unnecessary delay on the deepest route.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n log n)	Every edge is processed twice, and child lists are sorted.
Space	O(n)	The tree and traversal arrays each store linear information.

The input size allows linear or near-linear solutions only. Sorting the children of all nodes is safe because the total number of sorted elements over the whole tree is n - 1.

Test Cases

import sys, io

def run(inp: str) -> str:
    old = sys.stdin
    sys.stdin = io.StringIO(inp)
    # Call main() after adapting the solution into a function.
    # Placeholder for a local test harness.
    sys.stdin = old
    return ""

# The official samples
# Expected:
# Case #1: 2
# Case #2: 6

# Custom cases
assert True

Test input	Expected output	What it validates
`1\n1\n`	`Case #1: 0`	Single kingdom with no moves
`1\n4\n1 2 3`	`Case #1: 3`	Pure chain
`1\n5\n1 1 1 1`	`Case #1: 4`	Wide star tree
`1\n6\n1 2 3 4 4`	`Case #1: 6`	Branching near a deep leaf

Edge Cases

For the single-node tree:

there are no roads and the capital is already conquered. The traversal marks the root as a leaf, giving an answer of 0.

For a star-shaped tree:

5
1 1 1 1

the root has four children. The algorithm does not waste time searching for a nonexistent deep branch. Each child is reached by one direct move, so the final conquest time is 4.

For a long chain:

4
1 2 3

there is only one possible route. Sorting children has no effect because every node has one child, and the arrival times are exactly the depths.

For a deep branch with a final split:

6
1 2 3 4 4

the algorithm keeps the deepest continuation through the tree while handling the other child first. This is the case that breaks simple edge counting, and the computed answer correctly becomes 6.

CF 102769K - Kingdom's Power