338. Familystrokes Access

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

internalCnt ← 0 // |I| horizontalCnt ← 0 // # childCount(v) ≥ 2

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

long long internalCnt = 0; // import sys sys.setrecursionlimit(200000) 338. FamilyStrokes

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);

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.

Proof. By definition a leaf has no children, thus rule 1 (vertical stroke) and rule 2 (horizontal stroke) are both inapplicable. ∎ Every internal node (node with childCnt ≥ 1 ) requires exactly one vertical stroke . root = 1 stack = [(root, 0)] #

while stack not empty: v, p = pop(stack) childCnt = 0 for each w in G[v]: if w == p: continue // ignore the edge back to parent childCnt += 1 push (w, v) on stack

print(internal + horizontal)

def main() -> None: data = sys.stdin.read().strip().split() if not data: return it = iter(data) n = int(next(it)) g = [[] for _ in range(n + 1)] for _ in range(n - 1): u = int(next(it)); v = int(next(it)) g[u].append(v) g[v].append(u) ∎ answer = internalCnt + horizontalCnt computed by

const int ROOT = 1; vector<int> parent(N + 1, 0); vector<int> st; // explicit stack for DFS st.reserve(N); st.push_back(ROOT); parent[ROOT] = -1; // mark visited

Proof. The drawing rules require a vertical line from the node down to the row of its children whenever it has at least one child. The line is mandatory and unique, hence exactly one vertical stroke. ∎ An internal node requires a horizontal stroke iff childCnt ≥ 2 .