Overview

Two rooted trees (Takahashi’s facility and Aoki’s facility) are given. For each tree, we generate all possible “DFS traversal records” (appending a when entering and b when leaving) by choosing the visiting order of children arbitrarily, and then enumerate the common elements of these two sets in lexicographic order.

Since the number of nodes \(N\) is at most 10, which is very small, we can solve this with an approach that recursively generates all possible strings.

Analysis

Structure of Traversal Records

The traversal record for a subtree rooted at node \(v\) has the following form: 'a' + (record of child 1's subtree) + (record of child 2's subtree) + ... + (record of child k's subtree) + 'b'

Since the visiting order of children is arbitrary, there are as many candidate patterns as there are permutations of children \(1, \dots, k\). Furthermore, each child’s subtree can have multiple possible traversal records, so we also need to consider the combinations (Cartesian product) of those.

Eliminating Duplicates

When there are multiple subtrees with the same structure, swapping them produces the same string. To avoid unnecessary computation, it is efficient to sort and deduplicate the generated set of strings while processing recursively.

Algorithm

For each node \(u\), consider a function get_all_strings(u) that returns the set of all possible traversal records (a list of strings) that can be generated from its subtree.

1. Base case: If node \(u\) is a leaf (has no children), return {"ab"}.
2. Recursive step:
   • For all children \(v_1, v_2, \dots, v_k\) of \(u\), recursively call get_all_strings(v_i) to obtain the string set for each child.
   • Try all possible orderings of the children’s sets (using next_permutation).
   • For each ordering, generate all combinations (Cartesian product) of selecting one string from each child’s set, and concatenate them in the form 'a' + ... + 'b'.
   • Sort all obtained strings, remove duplicates, and return the result for \(u\).
3. Final processing:
   • Compute the set \(S_A\) obtained from Takahashi’s tree (root = 1) and the set \(S_B\) obtained from Aoki’s tree (root = 1).
   • Find the intersection of \(S_A\) and \(S_B\), and output its size and contents.

Complexity

Under the constraint \(N \le 10\), the total number of generated strings is not too large. - Time complexity: In the worst case, it is related to the total number of bracket sequences with \(N\) pairs (Catalan number \(C_{N-1}\)) and the combinations of permutations, but for \(N=10\) it runs sufficiently fast. - Space complexity: The lists of strings held at each node are dominant, but they also fit within memory limits.

Implementation Notes

• Generating permutations: When using std::next_permutation, by sorting the children’s sets themselves beforehand, duplicate permutations for children with the same structure (children that return the same string set) can be automatically skipped.

• Generating Cartesian products: After fixing the order of children, the process of selecting one string from each child’s set can be implemented using recursion, nested loops, or backtracking.

• Computing the intersection: Using std::set_intersection, common elements can be efficiently extracted from two sorted vectors.

  Source Code

#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <functional>
#include <iterator>

using namespace std;

/**
 * Problem: Common Traversal Records
 * Given two rooted trees with N nodes, both rooted at node 1.
 * A traversal record is generated by a DFS traversal where children are visited in any order.
 * 'a' is added when entering a node, and 'b' is added when leaving.
 * We need to find the intersection of all possible traversal records for both trees
 * and list them in lexicographical order.
 * N is small (N <= 10), so the number of possible strings is manageable (at most Catalan number C_{N-1}).
 */

// Function to generate all unique traversal strings for a subtree rooted at u
vector<string> get_all_strings(int u, int p, const vector<vector<int>>& adj) {
    vector<vector<string>> child_sets;
    for (int v : adj[u]) {
        if (v == p) continue;
        // Recursively get all possible traversal records for each child's subtree
        child_sets.push_back(get_all_strings(v, u, adj));
    }
    
    // Base case: If u is a leaf, the only traversal record is "ab" (entering then leaving)
    if (child_sets.empty()) {
        return {"ab"};
    }

    // Sort the sets of strings from children to handle identical children efficiently
    // This allows next_permutation to generate only unique orderings of the children.
    sort(child_sets.begin(), child_sets.end());
    
    vector<string> results;
    int k = child_sets.size();
    string current = "";

    // Recursive function to pick one string from each child's set and concatenate them
    function<void(int)> combine = [&](int idx) {
        if (idx == k) {
            // A traversal record for the current node is 'a' + concatenated child strings + 'b'
            results.push_back("a" + current + "b");
            return;
        }
        int old_size = current.size();
        for (const string& s : child_sets[idx]) {
            current += s;
            combine(idx + 1);
            current.resize(old_size); // Backtrack
        }
    };

    // next_permutation generates all unique permutations of the children's sets
    do {
        combine(0);
    } while (next_permutation(child_sets.begin(), child_sets.end()));

    // Sort and remove duplicates to return only unique traversal records for this subtree
    sort(results.begin(), results.end());
    results.erase(unique(results.begin(), results.end()), results.end());
    return results;
}

int main() {
    // Optimize standard I/O
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int N;
    if (!(cin >> N)) return 0;
    
    // Adjacency lists for Takahashi's facility (Tree A) and Aoki's facility (Tree B)
    vector<vector<int>> adjA(N + 1), adjB(N + 1);
    for (int i = 0; i < N - 1; ++i) {
        int u, v;
        if (!(cin >> u >> v)) break;
        adjA[u].push_back(v);
        adjA[v].push_back(u);
    }
    for (int i = 0; i < N - 1; ++i) {
        int u, v;
        if (!(cin >> u >> v)) break;
        adjB[u].push_back(v);
        adjB[v].push_back(u);
    }

    // Generate all possible traversal records for both facilities starting from root 1
    vector<string> setA = get_all_strings(1, 0, adjA);
    vector<string> setB = get_all_strings(1, 0, adjB);

    // Find the common records by intersecting the two sorted sets
    vector<string> common;
    set_intersection(setA.begin(), setA.end(), setB.begin(), setB.end(), back_inserter(common));

    // Output the total number of common records
    cout << common.size() << "\n";
    // Output each common record in lexicographical order
    for (const string& s : common) {
        cout << s << "\n";
    }

    return 0;
}

This editorial was generated by gemini-3-flash-thinking.