結果

問題 No.1928 Make a Binary Tree
ユーザー gew1fw
提出日時 2025-06-12 20:06:31
言語 PyPy3
(7.3.15)
結果
WA  
実行時間 -
コード長 2,258 bytes
コンパイル時間 396 ms
コンパイル使用メモリ 82,532 KB
実行使用メモリ 127,320 KB
最終ジャッジ日時 2025-06-12 20:13:30
合計ジャッジ時間 11,186 ms
ジャッジサーバーID
（参考情報） 		judge3 / judge4
このコードへのチャレンジ
（要ログイン）
ファイルパターン 結果
sample AC * 3
other AC * 4 WA * 29 TLE * 1 -- * 23
権限があれば一括ダウンロードができます

ソースコード

diff # 

from collections import deque
import sys

def main():
    sys.setrecursionlimit(1 << 25)
    n = int(sys.stdin.readline())
    if n == 1:
        print(1)
        return

    adj = [[] for _ in range(n + 1)]
    for _ in range(n - 1):
        x, y = map(int, sys.stdin.readline().split())
        adj[x].append(y)
        adj[y].append(x)

    parent = [0] * (n + 1)
    depth = [0] * (n + 1)
    visited = [False] * (n + 1)
    q = deque([1])
    visited[1] = True
    parent[1] = None

    while q:
        u = q.popleft()
        for v in adj[u]:
            if not visited[v] and v != parent[u]:
                parent[v] = u
                depth[v] = depth[u] + 1
                visited[v] = True
                q.append(v)

    # Sort nodes excluding root (1) in decreasing order of depth
    nodes = [i for i in range(2, n + 1)]
    nodes.sort(key=lambda x: -depth[x])

    up = [0] * (n + 1)
    for u in range(1, n + 1):
        up[u] = u
    children = [0] * (n + 1)
    count = 1  # root is always present

    for v in nodes:
        u = parent[v]
        assigned = False
        while u is not None:
            current_up = up[u]
            # Find the first available ancestor
            while current_up is not None and children[current_up] >= 2:
                # Move up to parent's up
                next_up = parent[current_up]
                if next_up is None:
                    up[u] = None
                    current_up = None
                else:
                    up[u] = up[next_up]
                    current_up = up[u]
            if current_up is not None and children[current_up] < 2:
                children[current_up] += 1
                if children[current_up] == 2:
                    # Update up[u] to the next available ancestor of current_up's parent
                    next_parent = parent[current_up]
                    if next_parent is None:
                        up[u] = None
                    else:
                        up[u] = up[next_parent]
                assigned = True
                break
            else:
                # Move up to parent of u
                u = parent[u]
        if assigned:
            count += 1

    print(count)

if __name__ == "__main__":
    main()
0