Maximum Squads Calculation

Assessing knight pairs

I'm working through forming pairs of knights without ancestor/descendant relationships. In a hierarchical structure, this means picking pairs where no knight is directly or indirectly superior to the other.

Setting pairing limits

Interestingly enough, each knight can only be in one squad. The king, who is an ancestor to all, is excluded from consideration since no knight can be superior to their squadmate.

Mapping out hierarchy

OK, I’m clarifying that the king's ultimate ancestry excludes him from the mating pool. Pairing requires knights to be in separate branches from any node's viewpoint, excluding their subtree nodes.

Crafting pairs

I'm digging into a strategy for maximizing knight pairs, avoiding ancestor-descendant links, and focusing on avoiding connections by selecting nodes from different branches of the tree.

Formulating pairs

I’m strategizing to root the tree at the King, ensuring nodes are chosen from different branches with respect to the King, while avoiding connecting nodes to their ancestors or descendants.

Laying out the options

I'm leaning towards a greedy algorithm, which avoids ancestor/descendant knight pairs by excluding parents or children, aiming for the maximum possible matches.

Thinking through constraints

I’m evaluating a bipartite graph or independent set to satisfy the constraint of selecting nodes without paths to or from their subtrees or ancestors.

Mapping relationships

I’m working through a DFS approach, ensuring each knight pair avoids any ancestor/descendant ties. Valid pairs have an LCA that isn't one or both knights.

Assessing connection limits

I’m mapping out that valid pairs must stem from different branches, ensuring neither knight is on the path to the root of the other.

Crafting valid matches

I'm piecing together pairs of knights, ensuring no ancestor/descendant linking. This leads me to consider children of the same node as a viable option.

Assessing different approaches

I’m thinking through dynamic programming, tracking node usage to ensure pairs don’t share ancestor/descendant relationships, and optimizing for the maximum number of valid pairs.

Evaluating combinatorial methods

I’m thinking about combinatorial strategies to maximize valid knight pairs, ensuring no ancestor/descendant relationships, and exploring node pairing within the same subtree with the same ancestor.

Exploring subtree dynamics

OK, let me see. Pairing knights from different subtrees of the king avoids ancestor/descendant issues, so grouping by subtree seems key for maximizing valid pairs.

Mapping the strategy

I’m organizing knights by subtree, counting the king's children, and forming pairs across different subtrees to maximize the number of valid knight pairs.

Unleashing recursive possibilities

I'm thinking through recursive strategies to form valid knight pairs across various subtrees, ensuring no ancestor/descendant links. This opens up numerous options for forming pairs.

Piecing it together

I'm piecing together that the maximum valid knight pairs in a rooted tree is the floor of n/2, considering node constraints and hierarchical structures.

Adjusting my approach

In a strictly hierarchical chain of length n, valid knight pairs are not possible due to ancestor/descendant relationships, resulting in zero valid pairs.

Piecing it together

Tree structure significantly impacts the number of valid knight pairs. A purely chain-like tree results in zero pairs, while a star structure with the king as root results in pairs equal to floor((n-1)/2).

Formulating a dynamic strategy

I’m thinking through a dynamic programming approach, combining node and subtree counts while avoiding ancestor relationships, using a "tree dp from the bottom up" method.

Piecing together

I’m thinking about pairing leaves to simplify the problem, as they have no ancestor relationships with each other outside their own subtree.

Pairing knights effectively

I'm thinking through how to pair knights based on their depth or level from the root, considering ancestor relationships and ensuring no superclass pairs.

Pairing nodes by depth

OK, let me see. If two nodes share the same depth, they can't be ancestor/descendant, which simplifies the process of forming valid knight pairs.

Pairing by depth

I’m figuring out that knights at the same depth can pair off, avoiding ancestor/descendant issues. Valid pairs come from within each depth level. Progress is steady and promising.

Evaluating pair potential

I’m mapping potential pairs by assessing if a node at depth d can be an ancestor of a node at depth e, considering paths to the root and ancestor relationships.

Assessing knight pairings

Pairing knights by depth can ignore unpaired knights from different depths. By examining combinations, we can refine the strategy to maximize pairings effectively.

Wondering about leftover knights

I’m thinking left unpaired knights from different depths might form extra valid pairs, but efficiently handling this for large depths is still up for debate.

Evaluating pairing strategies

I'm thinking through summing floor values of knights per depth to form pairs, then ensuring no ancestor/descendant relationships among leftover knights to maximize valid pairs.

Forming cross-depth pairs

I'm figuring out how leftover knights from different depths might pair up, considering the structure of the tree and avoiding ancestor/descendant relationships. This approach holds promise.

Mapping out subtree pairings

I’m thinking about forming pairs from leftover knights by examining subtree relationships and ensuring no ancestor/descendant connections between nodes, with BFS or DFS to gauge subtree sizes.

Mapping connections

OK, let me see. I'm piecing together the largest matching by identifying the most edges in the complement of the partial order graph, ensuring no ancestor/descendant relationships.

Piecing together max matching

I'm working through the comparability graph's complement, ensuring no ancestor/descendant connections. The largest matching is determined by the largest antichain's size via the floor function.

Considering tree depth

Mapping out a 2-color bipartite tree by depth parity, ensuring pairs are incomparable and no node is duplicated. This organization is crucial to tackle the problem.

Listening to connections

I’m considering whether nodes at even depths can form ancestor/descendant relationships. Paths with an odd number of steps, leading to nodes of different parities, are interesting to note.

Organizing nodes

I’m piecing together a simple maximum bipartite matching strategy: sort nodes by depth parity, link them if they’re not ancestor/descendant, and ensure each node is matched only once.

Pairing knights

I’m mapping out the strategy to pair knights from different subtrees, ensuring no ancestor/descendant relationships. This approach leverages tree-based dynamic programming to manage complexity efficiently.

Crafting maximum matches

I'm forming maximum matches across subtrees, treating them as distinct sets with nodes freely pairing with nodes from another set, akin to a complete bipartite graph.

Crafting recursive tree pairs

I’m thinking about a recursive method to pair nodes, grouping subtrees into a multi-partite graph, and focusing on balancing pairs across nodes and sub-subtrees for the best outcome.

Mapping out the strategy

It's interesting to see that keeping the root T unmatched might lead to the best results, allowing it to pair with nodes from other sibling subtrees.

Assessing options

I'm digging into simpler approaches, like focusing on pairs from different branches or levels, and using sample examples to gauge feasibility.

Identifying potential pairs

I’m thinking through a stable matching concept: start by pairing leaves from different subtrees, then remove them. Continue until no more viable matches are left.

Forming initial pairs

I’m pairing leaves from different subtrees, then removing them to iterate this process recursively. This approach ensures each node gets matched only once.

Forming valid leaf pairs

I’m noting that any two distinct leaves can be matched without ancestor/descendant relationships, and then removing them from the tree ensures each node is paired only once.

Ensuring valid leaf pairs

I’m making sure each leaf pair is valid, ensuring no ancestor/descendant relationship. Removing a leaf may convert its parent to a leaf, ready for the next iteration.

Ensuring valid matches

OK, let me see. Pairing leaves is straightforward—no ancestor/descendant issues. Internal nodes might need careful handling once they transform into leaves.

Greedily pairing leaves

I’m focusing on quickly pairing leaves from different subtrees, ensuring no ancestor/descendant relationships, then removing them from the queue. This method simplifies the problem effectively.

Crafting the pairing process

I’m thinking through making pairs from different subtrees, ensuring no ancestor/descendant ties. Removing paired leaves and their parents efficiently manages complexity.

Identifying leaf pairs

Pairing knights in a chain involves identifying the last non-leaf knight as a leaf. Then, I remove it and check for more pairs. Let me see if this properly pairs up everyone.

Validating leaf pairing

OK, let me see. I’m confirming that in a star setup, each leaf pairs without ancestor/descendant issues, consistently forming pairs up to floor(n/2). It’s a reliable method.

Forming pairs

OK, let me see. I’m thinking through an unbalanced tree, identifying leaves, and pairing them up while removing them from the queue to handle complexity efficiently.

Pairing knights

I’m noticing that pairing knights in an unbalanced tree involves identifying leaves, forming pairs, and removing them one by one, efficiently managing complexity and preventing ancestor/descendant relationships.

Forming valid pairs

I’m ensuring each pair picked works with no ancestor/descendant links. Noting that all knights have been matched, the solution is efficient and checks out.

Ensuring valid pairings

Confirming all knight pairings are ancestor/descendant-free, and every knight found a match. This approach highlights the validity and completeness of our matching strategy.

Crafting the code

I’m assembling the solution, focusing on creating the adjacency list, identifying leaves, and forming pairs without ancestor/descendant relationships. Each knight's parent-child connections are being meticulously mapped.

Laying out the options

I’m thinking through forming knight pairs by using adjacency lists, identifying leaves, and ensuring no ancestor/descendant ties, with a specific focus on when n=2.

Mapping out the solution

OK, let me see. I’m outlining the solution: constructing adjacency lists, identifying leaves, systematically forming pairs without ancestor/descendant relationships, and effectively managing complexity throughout.

Ensuring unique pushes

I’m making sure each leaf node in the queue is unique and accurately processed. This involves checking child counts and tracking visited nodes to avoid duplicates and confirm accurate pairings.

Formulating a plan

I’m confirming that the approach ensures each knight is paired only once by removing nodes from the tree once they’re used. This approach guarantees no knight is part of more than one squad.

Mapping the strategy

I'm detailing the solution: identifying leaves, forming valid pairs, and ensuring each knight is matched only once. This method effectively manages complexity for large-sized trees.

Thought about knight squad pairing for 1m 35s

Analysis

We have $n$ knights labeled from $0$ to $n-1$, where knight $0$ is the King. Each knight $i$ ($1 \le i \le n-1$) has a direct superior given by arr[i-1]. A knight $X$ is a (direct or indirect) superior of $Y$ if $X$ is on the chain of command from $Y$ up to the King.

We want to form as many two-knight squads as possible, subject to:

1. No knight can appear in more than one squad.
2. In a squad of two knights $(A, B)$, neither $A$ is superior of $B$, nor $B$ is superior of $A$.
   • Equivalently, in the tree rooted at the King, $A$ and $B$ must not lie on the same root-to-leaf path (i.e., they must not be in an ancestor–descendant relationship).

Because the King $(0)$ is an ancestor of everyone else, he can never pair with any knight (that would violate the “no superior” condition). So effectively, we only consider the sub-forest of knights $1,2,\ldots,n-1$ and ignore knight $0$ in the pairing.

A key insight is:

Any two leaves in this sub-forest can safely form a valid squad,
since a leaf cannot be an ancestor of any other leaf.

Hence, a greedy “peel the leaves” strategy works:

1. Identify all knights (among $1 \dots n-1$) who have no children in this sub-forest (i.e., outdegree $=0$). These are our current leaves.
2. Repeatedly match (pop) any two leaves at a time to form a squad.
   • Once two leaves $a,b$ are matched, remove them from the tree (so they can’t appear in any other squad).
   • Removing them may cause their respective parents to become new leaves (child-count goes down). If a parent’s child count becomes zero, it becomes a leaf—add it to the leaves-queue.
3. Continue until fewer than two leaves remain.

This procedure gives the maximum number of valid squads.

Why it works:

• Every time you pick two leaves, they definitely do not lie on an ancestor–descendant path (a leaf is never ancestor of another leaf).
• Greedily pairing leaves can never block a better solution: any node that could be paired “better” with an internal node will eventually become a leaf itself (once all its children are removed), so it is safe (and optimal) to keep pairing leaves as soon as they appear.
• In extreme examples:
  • A chain (like $0 \to 1 \to 2 \to 3 \to \cdots$) ends up producing 0 squads because there is effectively only one leaf at a time in the sub-forest.
  • A star (King $\to$ many direct children) pairs up nearly everyone ($\lfloor (n-1)/2 \rfloor$ squads).

This algorithm runs in $O(n)$ time (we only keep track of parents and child counts, then push/pop from a queue).

C++ Code

#include <bits/stdc++.h>
using namespace std;

/*
    Given an array 'arr' of size n-1, describing the direct superior for knights 1..n-1,
    where:
        - n = arr.size() + 1
        - arr[i-1] = parent of knight i
        - King is knight 0 and never pairs (he is an ancestor to all),
    return the maximum number of 2-knight squads we can form such that
    no knight is in more than one squad, and no two knights in a squad
    have a superior-subordinate (ancestor-descendant) relationship.
*/
int maximumPossibleSquads (vector<int> &arr) {
    int n = (int)arr.size() + 1;  // total knights = size of arr + 1 (including King)
    if(n <= 2) {
        // If n=2, we have knights: 0(King),1 => can't form any squad.
        return 0;
    }

    // parent[i] will store the parent of knight i in the sub-forest 1..n-1.
    // If a knight's parent is 0 (the King), we set parent[i] = -1 to mark "no parent in sub-forest".
    vector<int> parent(n, -1);

    // childCount[i] = number of children knight i has within 1..n-1
    // (We don't include the King as a child or parent in this sub-forest.)
    vector<int> childCount(n, 0);

    // Build parent/childCount for knights 1..n-1
    // i ranges 1..n-1, while arr[i-1] is the direct superior of i
    for(int i = 1; i < n; i++){
        int p = arr[i-1]; 
        if(p != 0) {
            // parent is a knight in [1..n-1]
            parent[i] = p;
            childCount[p]++;
        } else {
            // p == 0 => King is the parent => ignore in the sub-forest
            parent[i] = -1; 
        }
    }

    // All nodes (1..n-1) that have childCount == 0 are leaves in the sub-forest
    queue<int> leaves;
    for(int i = 1; i < n; i++){
        if(childCount[i] == 0){
            leaves.push(i);
        }
    }

    int squads = 0;  // count how many 2-knight squads we can form

    // While we can pop at least 2 leaves, form a squad
    while((int)leaves.size() >= 2) {
        int a = leaves.front(); leaves.pop();
        int b = leaves.front(); leaves.pop();

        // We have matched knights a and b
        squads++;

        // Remove them from the sub-forest
        // Decrement their parents' child counts
        int pa = parent[a];
        int pb = parent[b];
        if(pa != -1){
            childCount[pa]--;
            // if that parent becomes a leaf, push it
            if(childCount[pa] == 0){
                leaves.push(pa);
            }
        }
        if(pb != -1){
            childCount[pb]--;
            if(childCount[pb] == 0){
                leaves.push(pb);
            }
        }
    }

    return squads;
}

// A main() for quick tests, remove or comment out if not needed in the platform
int main(){
    // Example usage:
    // Suppose n=6 => arr.size()=5 => knights are 0..5, with 0=King
    // arr[i-1] = parent of knight i
    // Let's try an example where:
    //   arr = [0,1,2,0,0]
    //   - knight 1 -> parent=0 (King)
    //   - knight 2 -> parent=1
    //   - knight 3 -> parent=2
    //   - knight 4 -> parent=0 (King)
    //   - knight 5 -> parent=0 (King)
    // That means we have the tree:
    //    0(King)
    //    /   |   \
    //   1    4    5
    //   |
    //   2
    //   |
    //   3
    // Maximum squads? We'll see:

    vector<int> arr = {0,1,2,0,0}; 
    cout << maximumPossibleSquads(arr) << endl;
    // You can add more tests here.

    return 0;
}

Explanation of the Key Steps

1. Ignore the King: Knight $0$ cannot pair with anyone because he is an ancestor of all.
2. Build the sub-forest of knights $1$ through $n-1$. We store:
   • parent[i] = arr[i-1] if that parent $\neq 0$; otherwise $-1$ for “no parent in sub-forest.”
   • childCount[i] = number of children of knight $i$ in this sub-forest.
3. Initialize all leaves (where childCount[i] == 0) in a queue.
4. Greedy pairing of leaves:
   • Pop two leaves at a time to form a squad.
   • Removing a leaf reduces its parent's child count; if that parent becomes a new leaf, enqueue it.
5. Continue until we cannot pair anymore (queue has fewer than 2 leaves).

This correctly finds the maximum number of two-knight squads satisfying the “no ancestor–descendant” constraint and “no knight used more than once.”