Overview

For each tree, we enumerate all possible traversal records (a/b strings) that can be created by freely changing the order in which children are visited, and finally output the intersection of the two sets in lexicographic order.
Since the constraint \(N \le 10\) is small, recursively constructing the set of possible strings for each subtree is fast enough.

Analysis

The traversal record at each vertex \(v\) always has the form:

• a when entering
• Process children in order
• b when leaving

In other words, the record for vertex \(v\) is always:

\[ \text{"a"} + (\text{concatenation of child subtree records in order}) + \text{"b"} \]

Key Observation

• Since only the order of children is a degree of freedom, the candidate strings for vertex \(v\) can be constructed by combining “the candidate set for each child” × “permutations of children.”
• For a leaf, there is only one candidate: ab.
• The candidates for the entire tree are the candidate set of the root (room 1).

Issues with the Naive Approach

Directly trying all DFS orderings for the entire tree involves permutation branching at every vertex, making implementation complex.
Additionally, the same string can be generated multiple times (e.g., from isomorphic subtrees), so deduplication is necessary.

Solution

• First, root the tree at vertex 1 and build the children arrays.
• dfs(v) returns “the set of all possible traversal records from the subtree rooted at vertex \(v\).”
• Enumerate all permutations of children using next_permutation, and for each permutation, concatenate via Cartesian product.
• Insert into a set<string> for deduplication.
• Perform this separately for Takahashi’s tree and Aoki’s tree, then take the set intersection at the end.

Algorithm

1. Traverse the input undirected tree from root 1 using BFS, constructing parent and children.
2. Recursive function dfs(v):
   • For each child c, obtain childSets[c] = dfs(c).
   • If there are 0 children, return {"ab"}.
   • Iterate over all permutations of the child index array idx = [0,1,...,k-1].
   • For each permutation,
     starting with cur = {""}, successively compute cur = { pre + add | pre in cur, add in childSets[id] } to build concatenation candidates.
   • For each mid in cur, add "a" + mid + "b" to the set.
   • Return the contents of the set as a vector<string>.
3. Obtain the root candidate sets SA and SB for each of the two trees.
4. After sorting and deduplicating SA and SB, compute the intersection inter using set_intersection.
5. Output the count and the strings in order.

Complexity

• Time complexity: Strictly depends on the total number of generated strings.
  At each vertex, we enumerate “\(k!\) child permutations” and “the Cartesian product of child sets,” so
  the overall complexity is roughly proportional to the output size (and intermediate generation size), which is exponential.
  However, since \(N \le 10\), this is sufficiently feasible.
• Space complexity: Depends on the total size of generated and stored string sets (also dependent on output size).

Implementation Notes

• Converting to a rooted tree (assigning parent-child directions) first makes the recursion easier to write.

• Since the same string can be generated through multiple paths, use set<string> at each vertex for deduplication.

• Sorting SA and SB for the final intersection allows the use of set_intersection, keeping the code concise.

• The string length is always \(2N\), so comparisons (lexicographic order) are easy to handle.

  Source Code

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

static vector<string> enumerate_records(int N, const vector<vector<int>>& g) {
    vector<int> parent(N + 1, -1);
    vector<vector<int>> children(N + 1);

    // root at 1
    {
        queue<int> q;
        q.push(1);
        parent[1] = 0;
        while (!q.empty()) {
            int v = q.front(); q.pop();
            for (int to : g[v]) {
                if (parent[to] != -1) continue;
                parent[to] = v;
                children[v].push_back(to);
                q.push(to);
            }
        }
    }

    function<vector<string>(int)> dfs = [&](int v) -> vector<string> {
        vector<vector<string>> childSets;
        childSets.reserve(children[v].size());
        for (int c : children[v]) childSets.push_back(dfs(c));

        int k = (int)children[v].size();
        if (k == 0) return vector<string>{"ab"};

        vector<int> idx(k);
        iota(idx.begin(), idx.end(), 0);

        set<string> uniq; // deduplicate due to identical child-subtrees' strings

        do {
            vector<string> cur{""};
            for (int id : idx) {
                vector<string> nxt;
                nxt.reserve(cur.size() * childSets[id].size());
                for (const string& pre : cur) {
                    for (const string& add : childSets[id]) {
                        nxt.push_back(pre + add);
                    }
                }
                cur.swap(nxt);
            }
            for (auto &mid : cur) {
                uniq.insert("a" + mid + "b");
            }
        } while (next_permutation(idx.begin(), idx.end()));

        vector<string> res(uniq.begin(), uniq.end());
        return res;
    };

    return dfs(1);
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int N;
    cin >> N;

    vector<vector<int>> gA(N + 1), gB(N + 1);
    for (int i = 0; i < N - 1; i++) {
        int u, v;
        cin >> u >> v;
        gA[u].push_back(v);
        gA[v].push_back(u);
    }
    for (int i = 0; i < N - 1; i++) {
        int u, v;
        cin >> u >> v;
        gB[u].push_back(v);
        gB[v].push_back(u);
    }

    vector<string> SA = enumerate_records(N, gA);
    vector<string> SB = enumerate_records(N, gB);

    vector<string> inter;
    {
        sort(SA.begin(), SA.end());
        SA.erase(unique(SA.begin(), SA.end()), SA.end());
        sort(SB.begin(), SB.end());
        SB.erase(unique(SB.begin(), SB.end()), SB.end());
        set_intersection(SA.begin(), SA.end(), SB.begin(), SB.end(), back_inserter(inter));
    }

    cout << inter.size() << '\n';
    for (auto &s : inter) cout << s << '\n';

    return 0;
}

---

This editorial was generated by gpt-5.3-codex.