**Subtask:** The initial and final trees have the same number of leaves.

Note that this condition implies that the set of leaves in the initial tree equals the set of leaves in the final tree.

Say that two nodes in the initial tree are in the same component if they end up being merged into the same node in the final tree. For this subtask only, the components may be uniquely determined. Note the following two rules:

- If $a$ is present in the final tree and is not the root, the parents of $a$ in the initial and final trees are in the same component.
- If $a$ and $b$ are in the same component, then the parents of $a$ and $b$ in the initial tree are in the same component.

More concisely,

par2(a) != 0 -> same_comp(par1(a), par2(a)) same_comp(a, b) -> same_comp(par1(a), par1(b))

We can figure out which nodes are in the same components by first applying all rules of the first type, and then applying all rules of the second type while iterating over the nodes in decreasing order of depth. After this, we can iterate over the tree in increasing order of depth to merge all nodes in the same component into one. The time complexity is $\mathcal O(N)$, or $\mathcal O(N\alpha(N))$ if DSU is used.

Ben's code:

#include <bits/stdc++.h> using namespace std; struct DSU { vector<int> e; void init(int N) { e = vector<int>(N, -1); } int get(int x) { return e[x] < 0 ? x : e[x] = get(e[x]); } void unite(int x, int y) { // min(x,y) now points to max(x,y) x = get(x), y = get(y); if (x == y) return; if (x > y) swap(x, y); e[x] = y; } }; void dfs1(vector<vector<int>> &with_depth, const vector<vector<int>> &child, int x, int d) { with_depth[d].push_back(x); for (int c : child.at(x)) dfs1(with_depth, child, c, d + 1); } void solve() { int N; cin >> N; DSU D; D.init(N + 1); vector<int> par1(N + 1); vector<vector<int>> child1(N + 1); for (int i = 1; i < N; ++i) { int v, p; cin >> v >> p; par1[v] = p; child1[p].push_back(v); } int root = 1; while (par1[root]) ++root; vector<vector<int>> with_depth(N); dfs1(with_depth, child1, root, 0); int M; cin >> M; for (int i = 1; i < M; ++i) { int v, p; cin >> v >> p; D.unite(p, par1[v]); // type 1 } for (int d = N - 1; d >= 0; --d) // type 2 for (int x : with_depth[d]) D.unite(par1[x], par1[D.get(x)]); cout << N - M << "\n"; for (int d = 1; d < N; ++d) { for (int x : with_depth[d]) if (D.get(x) != x) cout << x << " " << D.get(x) << "\n"; } } int main() { ios::sync_with_stdio(false); cin.tie(nullptr); int T; cin >> T; while (T--) solve(); }

**Full Credit:**

Call a node *extant* if its value is in the final tree. We will calculate a
DP $dp(a, b)$ with boolean value equal to whether node $a$ can validly merge
into node $b$ by the end; that is, $dp(a, b)$ is true iff $b$ is extant and, if
at some point $a$ and $b$ are children of the same node, we can merge $a$ into
$b$ and be able to successfully obtain the final tree we desire.

In order for $dp(a, b)$ to be true, we need the following conditions to be satisfied:

- $b$ is extant
- The value of $a$ should be at most the value of $b$; otherwise, the value of $b$ would disappear even though it appears in the final tree (because $b$ is extant)
- If $a$ is extant, $a = b$
- For each child $c$ of $a$ in the original tree, there should exist a child $d$ of $b$ in the final tree such that $dp(c, d)$ is true; otherwise, we won't be able to merge $c$ in a way to end up with the final tree

$$\sum_a \sum_b (\text{# of children of $a$}) \cdot (\text{# of children of $b$}).$$

Since each node is the child of at most one other node, this is
$\mathcal O(N^2)$.
Given the DP, we can choose for each node $a$ what extant node it is merged into by iterating through nodes $a$ in increasing order of depth, and, given that $a$'s parent was merged into the extant node $b$, checking through each child $c$ of $b$ and choosing any $c$ such that $dp(a, c)$ is true, because that means that it is valid to merge $a$ into $c$. We can then output the actual merge of $a$ into $c$ at the same time (assuming $a \neq c$), because we will already have merged $a$'s parent into $b$ and so $a$ and $c$ will currently be siblings.

The DP computation step is $\mathcal O(N^2)$. The computation of merges is also $\mathcal O(N^2)$ because for each node we loop through the children of another node. Therefore, the algorithm is $\mathcal O(N^2)$ overall.

Danny's code (note that this code is actually $\mathcal O(N^3)$ because it searches through all nodes to find children):

import java.util.Scanner; public class TreeMerging { public static void main(String[] args) { Scanner in = new Scanner(System.in); for (int t = in.nextInt(); t > 0; t--) { int n = in.nextInt(); int root = (n * (n + 1)) / 2; int[] oldParent = new int[n + 1]; for (int edges = n - 1; edges > 0; edges--) { int a = in.nextInt(); int p = in.nextInt(); oldParent[a] = p; root -= a; } int m = in.nextInt(); int[] newParent = new int[n + 1]; boolean[] extant = new boolean[n + 1]; extant[root] = true; for (int edges = m - 1; edges > 0; edges--) { int a = in.nextInt(); int p = in.nextInt(); newParent[a] = p; extant[a] = true; } int[] depth = new int[n + 1]; for (int iteration = n; iteration > 0; iteration--) { for (int a = 1; a <= n; a++) { if (a != root) { depth[a] = depth[oldParent[a]] + 1; } } } boolean[][] canMerge = new boolean[n + 1][n + 1]; for (int d = n; d > 0; d--) { for (int a = 1; a <= n; a++) { if (depth[a] == d) { if (extant[a]) { canMerge[a][a] = true; } else { for (int b = a; b <= n; b++) { if (extant[b]) { canMerge[a][b] = true; for (int c = 1; c <= n; c++) { if (oldParent[c] == a) { boolean works = false; for (int cn = 1; cn <= n; cn++) { if (newParent[cn] == b && canMerge[c][cn]) { works = true; } } canMerge[a][b] &= works; } } } } } } } } System.out.println(n - m); int[] representative = new int[n + 1]; representative[root] = root; for (int d = 1; d <= n; d++) { for (int a = 1; a <= n; a++) { if (depth[a] == d) { for (int b = 1; b <= n; b++) { if (newParent[b] == representative[oldParent[a]] && canMerge[a][b]) { representative[a] = b; } } if (representative[a] != a) { System.out.println(a + " " + representative[a]); } } } } } } }