338. Familystrokes <2027>

Memory – The adjacency list stores 2·(N‑1) integers, plus a stack/queue of at most N entries and a few counters: O(N) .

internalCnt ← 0 // |I| horizontalCnt ← 0 // # v 338. FamilyStrokes

Only‑if childCnt = 1 : the sole child is placed directly under the parent; the horizontal segment would have length zero and is omitted by the drawing convention. ∎ The number of strokes contributed by a node v is Memory – The adjacency list stores 2·(N‑1) integers,

1 if childCnt(v) = 1 2 if childCnt(v) ≥ 2 0 if childCnt(v) = 0 Proof. Directly from Lemma 2 (vertical) and Lemma 3 (horizontal). ∎ answer = internalCnt + horizontalCnt computed by the algorithm equals the minimum number of strokes needed to draw the whole tree. Directly from Lemma 2 (vertical) and Lemma 3 (horizontal)

while stack: v, p = stack.pop() child_cnt = 0 for w in g[v]: if w == p: continue child_cnt += 1 stack.append((w, v)) if child_cnt: internal += 1 if child_cnt >= 2: horizontal += 1

int main() ios::sync_with_stdio(false); cin.tie(nullptr); int N; if (!(cin >> N)) return 0; vector<vector<int>> g(N + 1); for (int i = 0, u, v; i < N - 1; ++i) cin >> u >> v; g[u].push_back(v); g[v].push_back(u);

root = 1 stack = [(root, 0)] # (node, parent) internal = 0 horizontal = 0