Overview

This problem asks us to find, for each of two rooted trees, the set of all “traversal record” strings that can be generated by varying the order in which children are visited, and then output their intersection in lexicographic order. We solve this by full enumeration, leveraging the small constraint \(N \leq 10\).

Analysis

What is a Traversal Record?

A traversal record is the process of a DFS (depth-first search) on a tree recorded as a parenthesis-like string. When entering a node, we append a, and when leaving, we append b.

For example, if the root has two children, each of which is a leaf: - Enter root → a - Enter child 1 → a, leave child 1 → b - Enter child 2 → a, leave child 2 → b - Leave root → b

Result: not aababbb but a ab ab b = aababb (length \(2N = 6\))

Key Observations

• Changing the order in which children are visited can produce different traversal records.
• The string generated by each child’s subtree itself can have multiple patterns depending on the internal visitation order.
• Since \(N \leq 10\), the total number of generated strings is at most around \(9! = 362880\) (in the worst case of a star-shaped tree), making full enumeration well within time limits.

Approach

Since the constraints are very small, a brute-force approach that enumerates all possible traversal records for each tree and then computes the intersection of the two sets is effective.

Algorithm

1. Tree Construction: From the input graph, determine the parent-child relationships rooted at room 1 using BFS.

2. Full Enumeration of Traversal Records (recursive function generate):

   • For a leaf node: return only "ab"
   • For a node with children:
     1. Recursively compute the set of possible strings generated by each child’s subtree.
     2. Enumerate all combinations of choosing one string from each child.
     3. For each combination, try all permutations of the children’s order and add "a" + (concatenation of children's strings in that order) + "b" to the set.

3. Computing the Intersection: Find the set intersection of the two sets (set_intersection).

4. Output: Since we use set<string>, the results are automatically sorted in lexicographic order.

Complexity

• Time complexity: In the worst case, this is the product of the number of strings generated by each node’s children, multiplied by the number of permutations of the children. For \(N = 10\) with a star shape (root with 9 children), this gives \(9! = 362880\) patterns. Even when subtrees branch further, since \(N\) is small, this is sufficiently fast in practice. The strict upper bound is approximately \(O(N!)\).
• Space complexity: \(O(N! \cdot N)\) (to store the set of generated strings)

Implementation Notes

• Building the tree with BFS: Since the input is given as an undirected graph, we perform BFS from the root to determine parent-child relationships.

• Recursive set generation: A two-stage enumeration is required: first taking the “Cartesian product” of the string sets returned by each child’s subtree, then rearranging with permutations.

• Duplicate elimination: By using set<string>, duplicates that arise when different visitation orders produce the same string are automatically removed.

• Using set_intersection: The STL’s set_intersection can efficiently compute the intersection of sorted sets.

  Source Code

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

int N;
vector<int> adj[11];
vector<int> ch[11];

void buildTree(vector<int> adjs[], vector<int> chs[]) {
    for (int i = 1; i <= N; i++) chs[i].clear();
    vector<bool> visited(N + 1, false);
    queue<int> q;
    q.push(1);
    visited[1] = true;
    while (!q.empty()) {
        int u = q.front(); q.pop();
        for (int v : adjs[u]) {
            if (!visited[v]) {
                visited[v] = true;
                chs[u].push_back(v);
                q.push(v);
            }
        }
    }
}

set<string> generate(vector<int> chs[], int node) {
    if (chs[node].empty()) {
        return {"ab"};
    }
    
    int k = chs[node].size();
    vector<vector<string>> childStrings(k);
    for (int i = 0; i < k; i++) {
        auto s = generate(chs, chs[node][i]);
        childStrings[i] = vector<string>(s.begin(), s.end());
    }
    
    // Generate all combinations of one string from each child
    vector<vector<string>> allCombos;
    vector<string> cur(k);
    function<void(int)> genComb = [&](int idx) {
        if (idx == k) {
            allCombos.push_back(cur);
            return;
        }
        for (const string& s : childStrings[idx]) {
            cur[idx] = s;
            genComb(idx + 1);
        }
    };
    genComb(0);
    
    set<string> result;
    vector<int> perm(k);
    iota(perm.begin(), perm.end(), 0);
    
    for (const auto& combo : allCombos) {
        vector<int> p(k);
        iota(p.begin(), p.end(), 0);
        do {
            string s = "a";
            for (int i = 0; i < k; i++) {
                s += combo[p[i]];
            }
            s += "b";
            result.insert(s);
        } while (next_permutation(p.begin(), p.end()));
    }
    
    return result;
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    
    cin >> N;
    
    vector<int> adjA[11], adjB[11];
    vector<int> chA[11], chB[11];
    
    for (int i = 0; i < N - 1; i++) {
        int u, v;
        cin >> u >> v;
        adjA[u].push_back(v);
        adjA[v].push_back(u);
    }
    for (int i = 0; i < N - 1; i++) {
        int u, v;
        cin >> u >> v;
        adjB[u].push_back(v);
        adjB[v].push_back(u);
    }
    
    // Build trees rooted at 1
    // For A
    {
        vector<bool> visited(N + 1, false);
        queue<int> q;
        q.push(1); visited[1] = true;
        while (!q.empty()) {
            int u = q.front(); q.pop();
            for (int v : adjA[u]) {
                if (!visited[v]) {
                    visited[v] = true;
                    chA[u].push_back(v);
                    q.push(v);
                }
            }
        }
    }
    // For B
    {
        vector<bool> visited(N + 1, false);
        queue<int> q;
        q.push(1); visited[1] = true;
        while (!q.empty()) {
            int u = q.front(); q.pop();
            for (int v : adjB[u]) {
                if (!visited[v]) {
                    visited[v] = true;
                    chB[u].push_back(v);
                    q.push(v);
                }
            }
        }
    }
    
    set<string> setA = generate(chA, 1);
    set<string> setB = generate(chB, 1);
    
    vector<string> intersection;
    set_intersection(setA.begin(), setA.end(), setB.begin(), setB.end(), back_inserter(intersection));
    
    cout << intersection.size() << "\n";
    for (const string& s : intersection) {
        cout << s << "\n";
    }
    
    return 0;
}

This editorial was generated by claude4.6opus-thinking.