結果

問題 No.1928 Make a Binary Tree
コンテスト
ユーザー gew1fw
提出日時 2025-06-12 19:57:14
言語 PyPy3
(7.3.15)
結果
WA  
実行時間 -
コード長 2,022 bytes
コンパイル時間 348 ms
コンパイル使用メモリ 82,940 KB
実行使用メモリ 848,652 KB
最終ジャッジ日時 2025-06-12 19:58:49
合計ジャッジ時間 24,360 ms
ジャッジサーバーID
（参考情報） 		judge5 / judge3
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ファイルパターン 結果
sample AC * 3
other AC * 3 WA * 32 MLE * 12 -- * 10
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 # 

import sys
from collections import deque

def main():
    sys.setrecursionlimit(1 << 25)
    N = int(sys.stdin.readline())
    edges = [[] for _ in range(N+1)]
    for _ in range(N-1):
        x, y = map(int, sys.stdin.readline().split())
        edges[x].append(y)
        edges[y].append(x)
    
    # Build parent and children arrays using BFS
    parent = [0] * (N + 1)
    children = [[] for _ in range(N + 1)]
    visited = [False] * (N + 1)
    q = deque()
    q.append(1)
    visited[1] = True
    parent[1] = 0
    while q:
        u = q.popleft()
        for v in edges[u]:
            if not visited[v]:
                visited[v] = True
                parent[v] = u
                children[u].append(v)
                q.append(v)
    
    # Post-order traversal using iterative approach
    post_order = []
    stack = [(1, False)]
    while stack:
        node, processed = stack.pop()
        if processed:
            post_order.append(node)
            continue
        stack.append((node, True))
        # Push children in reverse order to process them left to right
        for c in reversed(children[node]):
            stack.append((c, False))
    
    # DP arrays
    dp = [0] * (N + 1)
    max1 = [0] * (N + 1)  # Top candidate value for each node
    max2 = [0] * (N + 1)  # Second top candidate value
    
    for u in post_order:
        candidates = []
        # Step 1: Add top two from each child's candidates
        for c in children[u]:
            candidates.append(max1[c])
            candidates.append(max2[c])
        # Step 2: Add DP of original children
        for v in children[u]:
            candidates.append(dp[v])
        
        # Find top two values in candidates
        m1 = m2 = 0
        for val in candidates:
            if val > m1:
                m2 = m1
                m1 = val
            elif val > m2:
                m2 = val
        max1[u] = m1
        max2[u] = m2
        dp[u] = 1 + m1 + m2
    
    print(dp[1])

if __name__ == "__main__":
    main()
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