#1129705 (PyPy3) No.3348 Tree Balance

• 問題: No.3348 Tree Balance
• ユーザー: ZOI-dayo
• 提出日時: 2025-10-29 16:22:52
• 言語: PyPy3 (7.3.15)
• 結果: MLE
• 実行時間: -
• コード長: 8,383 bytes
• コンパイル時間: 195 ms
• コンパイル使用メモリ: 82,220 KB
• 実行使用メモリ: 512,752 KB
• 最終ジャッジ日時: 2025-11-13 21:03:57
• 合計ジャッジ時間: 11,659 ms
• ジャッジサーバーID （参考情報）: judge2 / judge4

ファイルパターン	結果
other	RE * 24 MLE * 1

ソースコード

# https://github.com/tatyam-prime/SortedSet/blob/main/SortedSet.py
import math
from bisect import bisect_left, bisect_right
from typing import Generic, Iterable, Iterator, TypeVar
T = TypeVar('T')

class SortedSet(Generic[T]):
    BUCKET_RATIO = 16
    SPLIT_RATIO = 24
    
    def __init__(self, a: Iterable[T] = []) -> None:
        "Make a new SortedSet from iterable. / O(N) if sorted and unique / O(N log N)"
        a = list(a)
        n = len(a)
        if any(a[i] > a[i + 1] for i in range(n - 1)):
            a.sort()
        if any(a[i] >= a[i + 1] for i in range(n - 1)):
            a, b = [], a
            for x in b:
                if not a or a[-1] != x:
                    a.append(x)
        n = self.size = len(a)
        num_bucket = int(math.ceil(math.sqrt(n / self.BUCKET_RATIO)))
        self.a = [a[n * i // num_bucket : n * (i + 1) // num_bucket] for i in range(num_bucket)]

    def __iter__(self) -> Iterator[T]:
        for i in self.a:
            for j in i: yield j

    def __reversed__(self) -> Iterator[T]:
        for i in reversed(self.a):
            for j in reversed(i): yield j
    
    def __eq__(self, other) -> bool:
        return list(self) == list(other)
    
    def __len__(self) -> int:
        return self.size
    
    def __repr__(self) -> str:
        return "SortedSet" + str(self.a)
    
    def __str__(self) -> str:
        s = str(list(self))
        return "{" + s[1 : len(s) - 1] + "}"

    def _position(self, x: T) -> tuple[list[T], int, int]:
        "return the bucket, index of the bucket and position in which x should be. self must not be empty."
        for i, a in enumerate(self.a):
            if x <= a[-1]: break
        return (a, i, bisect_left(a, x))

    def __contains__(self, x: T) -> bool:
        if self.size == 0: return False
        a, _, i = self._position(x)
        return i != len(a) and a[i] == x

    def add(self, x: T) -> bool:
        "Add an element and return True if added. / O(√N)"
        if self.size == 0:
            self.a = [[x]]
            self.size = 1
            return True
        a, b, i = self._position(x)
        if i != len(a) and a[i] == x: return False
        a.insert(i, x)
        self.size += 1
        if len(a) > len(self.a) * self.SPLIT_RATIO:
            mid = len(a) >> 1
            self.a[b:b+1] = [a[:mid], a[mid:]]
        return True
    
    def _pop(self, a: list[T], b: int, i: int) -> T:
        ans = a.pop(i)
        self.size -= 1
        if not a: del self.a[b]
        return ans

    def discard(self, x: T) -> bool:
        "Remove an element and return True if removed. / O(√N)"
        if self.size == 0: return False
        a, b, i = self._position(x)
        if i == len(a) or a[i] != x: return False
        self._pop(a, b, i)
        return True
    
    def lt(self, x: T) -> T | None:
        "Find the largest element < x, or None if it doesn't exist."
        for a in reversed(self.a):
            if a[0] < x:
                return a[bisect_left(a, x) - 1]

    def le(self, x: T) -> T | None:
        "Find the largest element <= x, or None if it doesn't exist."
        for a in reversed(self.a):
            if a[0] <= x:
                return a[bisect_right(a, x) - 1]

    def gt(self, x: T) -> T | None:
        "Find the smallest element > x, or None if it doesn't exist."
        for a in self.a:
            if a[-1] > x:
                return a[bisect_right(a, x)]

    def ge(self, x: T) -> T | None:
        "Find the smallest element >= x, or None if it doesn't exist."
        for a in self.a:
            if a[-1] >= x:
                return a[bisect_left(a, x)]
    
    def __getitem__(self, i: int) -> T:
        "Return the i-th element."
        if i < 0:
            for a in reversed(self.a):
                i += len(a)
                if i >= 0: return a[i]
        else:
            for a in self.a:
                if i < len(a): return a[i]
                i -= len(a)
        raise IndexError
    
    def pop(self, i: int = -1) -> T:
        "Pop and return the i-th element."
        if i < 0:
            for b, a in enumerate(reversed(self.a)):
                i += len(a)
                if i >= 0: return self._pop(a, ~b, i)
        else:
            for b, a in enumerate(self.a):
                if i < len(a): return self._pop(a, b, i)
                i -= len(a)
        raise IndexError
    
    def index(self, x: T) -> int:
        "Count the number of elements < x."
        ans = 0
        for a in self.a:
            if a[-1] >= x:
                return ans + bisect_left(a, x)
            ans += len(a)
        return ans

    def index_right(self, x: T) -> int:
        "Count the number of elements <= x."
        ans = 0
        for a in self.a:
            if a[-1] > x:
                return ans + bisect_right(a, x)
            ans += len(a)
        return ans

import sys

# Pythonの再帰深度を増やす
sys.setrecursionlimit(200010) 

def main():
    # C++の高速化IOに相当
    
    try:
        # Nの読み込み
        n_line = sys.stdin.readline()
        if not n_line:
            return
        n = int(n_line)
        
        # wの読み込み
        w = list(map(int, sys.stdin.readline().split()))
        
        # グラフの構築 (隣接リスト)
        graph = [[] for _ in range(n)]
        for _ in range(n - 1):
            line = sys.stdin.readline().split()
            a, b = int(line[0]) - 1, int(line[1]) - 1 
            graph[a].append(b)
            graph[b].append(a)

    except (IOError, ValueError, EOFError):
        return

    # sum_subtree[i]: 頂点iを根とする部分木の重みの合計
    sum_subtree = list(w)
    
    # === dfs1: 各部分木の重みの合計を計算 ===
    def dfs1(i, p):
        for c in graph[i]:
            if c == p:
                continue
            sum_subtree[i] += dfs1(c, i)
        return sum_subtree[i]

    dfs1(0, -1)

    total = sum_subtree[0]
    
    # 答えを保持する (C++の参照渡しを模倣)
    ans = [float('inf')] 

    # === update関数 (C++のラムダ式update) ===
    def update(s1, s2):
        s3 = total - s1 - s2
        mn = min(s1, s2, s3)
        mx = max(s1, s2, s3)
        ans[0] = min(ans[0], mx - mn)

    # === dfs2: Small-to-Large (Sack) テクニック ===
    # C++の std::set* の代わりに、SortedSet オブジェクトを返す
    def dfs2(i, p):
        # 2. C++の `new set<ll>()` -> `SortedSet()`
        # これで O(log N) の操作が可能になる
        sums = SortedSet() 
        
        current = sum_subtree[i]
        remain = total - current

        for c in graph[i]:
            if c == p:
                continue
            
            res = dfs2(c, i) # res も SortedSet

            # --- 1. ネストしたカットの処理 ---
            # 3. C++の `res->lower_bound()` -> `res.bisect_left()`
            #    O(log |res|) で動作
            idx = res.bisect_left(current / 2)
            
            if idx < len(res):
                # 4. C++の `*itr` -> `res[idx]` (SortedSetはO(log N)で添字アクセス可能)
                update(remain, res[idx])
            if idx > 0:
                update(remain, res[idx - 1])

            # --- 2. Small-to-Large マージの準備 ---
            if len(sums) < len(res):
                sums, res = res, sums # 参照の交換 (O(1))
            
            # --- 3. 独立したカットの処理 ---
            for e in res: # O(|res|)
                # 5. `sums.bisect_left()` で O(log |sums|) の検索
                idx_s = sums.bisect_left((total - e) / 2)
                
                if idx_s < len(sums):
                    update(e, sums[idx_s]) # O(log |sums|)
                if idx_s > 0:
                    update(e, sums[idx_s - 1]) # O(log |sums|)
            
            # --- 4. マージ ---
            # 6. C++の `sums->insert(all(*res))`
            #    O(|res| * log |sums|)
            #    Pythonの SortedSet.update() も同様の計算量
            sums.update(res)

        # 7. C++の `sums->insert(current)` -> `sums.add(current)`
        #    O(log |sums|) で挿入
        sums.add(current)
        
        return sums

    # 根を0としてdfs2を実行
    dfs2(0, -1)

    # 最終的な答えの出力
    print(ans[0])

if __name__ == "__main__":
    main()