結果
問題 | No.2740 Old Maid |
ユーザー |
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提出日時 | 2024-04-20 16:41:35 |
言語 | PyPy3 (7.3.15) |
結果 |
AC
|
実行時間 | 694 ms / 2,000 ms |
コード長 | 5,067 bytes |
コンパイル時間 | 179 ms |
コンパイル使用メモリ | 82,564 KB |
実行使用メモリ | 155,544 KB |
最終ジャッジ日時 | 2024-10-12 12:30:39 |
合計ジャッジ時間 | 23,474 ms |
ジャッジサーバーID (参考情報) |
judge2 / judge4 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 62 |
ソースコード
import mathfrom bisect import bisect_left, bisect_rightfrom typing import Generic, Iterable, Iterator, List, Tuple, TypeVar, OptionalT = TypeVar('T')class SortedMultiset(Generic[T]):BUCKET_RATIO = 16SPLIT_RATIO = 24def __init__(self, a: Iterable[T] = []) -> None:"Make a new SortedMultiset from iterable. / O(N) if sorted / O(N log N)"a = list(a)n = self.size = len(a)if any(a[i] > a[i + 1] for i in range(n - 1)):a.sort()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 jdef __reversed__(self) -> Iterator[T]:for i in reversed(self.a):for j in reversed(i): yield jdef __eq__(self, other) -> bool:return list(self) == list(other)def __len__(self) -> int:return self.sizedef __repr__(self) -> str:return "SortedMultiset" + 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]: breakreturn (a, i, bisect_left(a, x))def __contains__(self, x: T) -> bool:if self.size == 0: return Falsea, _, i = self._position(x)return i != len(a) and a[i] == xdef count(self, x: T) -> int:"Count the number of x."return self.index_right(x) - self.index(x)def add(self, x: T) -> None:"Add an element. / O(√N)"if self.size == 0:self.a = [[x]]self.size = 1returna, b, i = self._position(x)a.insert(i, x)self.size += 1if len(a) > len(self.a) * self.SPLIT_RATIO:mid = len(a) >> 1self.a[b:b+1] = [a[:mid], a[mid:]]def _pop(self, a: List[T], b: int, i: int) -> T:ans = a.pop(i)self.size -= 1if not a: del self.a[b]return ansdef discard(self, x: T) -> bool:"Remove an element and return True if removed. / O(√N)"if self.size == 0: return Falsea, b, i = self._position(x)if i == len(a) or a[i] != x: return Falseself._pop(a, b, i)return Truedef lt(self, x: T) -> Optional[T]:"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) -> Optional[T]:"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) -> Optional[T]:"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) -> Optional[T]:"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 IndexErrordef 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 IndexErrordef index(self, x: T) -> int:"Count the number of elements < x."ans = 0for a in self.a:if a[-1] >= x:return ans + bisect_left(a, x)ans += len(a)return ansdef index_right(self, x: T) -> int:"Count the number of elements <= x."ans = 0for a in self.a:if a[-1] > x:return ans + bisect_right(a, x)ans += len(a)return ansN = int(input())P = list(map(int,input().split()))d = {}for i in range(N):d[P[i]] = iS = SortedMultiset([i for i in range(N)])SS = SortedMultiset([i + 1 for i in range(N)])ans = []while N:n = SS[0]if d[n] == S[-1]:n = SS[1]j = d[n]k = S.gt(j)S.discard(j)S.discard(k)ans.append(P[j])ans.append(P[k])SS.discard(P[j])SS.discard(P[k])N -= 2print(*ans)