338. Familystrokes Apr 2026

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

if childCnt > 0: // v has at least one child → internal internalCnt += 1 if childCnt >= 2: horizontalCnt += 1

print(internal + horizontal)

internalCnt ← 0 // |I| horizontalCnt ← 0 // # v

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); 338. FamilyStrokes

Both bounds comfortably meet the limits for N ≤ 10⁵ . Below are clean, self‑contained implementations in C++17 and Python 3 that follow the algorithm exactly. 6.1 C++17 #include <bits/stdc++.h> using namespace std;

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 . while stack: v, p = stack

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

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 By definition a leaf has no children, thus

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