結果

問題 No.2665 Minimize Inversions of Deque
ユーザー miya145592miya145592
提出日時 2024-03-08 22:57:50
言語 PyPy3
(7.3.15)
結果
AC  
実行時間 355 ms / 2,000 ms
コード長 1,988 bytes
コンパイル時間 177 ms
コンパイル使用メモリ 82,816 KB
実行使用メモリ 109,528 KB
最終ジャッジ日時 2024-09-29 20:28:00
合計ジャッジ時間 11,312 ms
ジャッジサーバーID
(参考情報)
judge3 / judge5
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ファイルパターン 結果
other AC * 40
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 #
プレゼンテーションモードにする

class SegTree:
def __init__(self, op, e, n, v=None):
self._n = n
self._op = op
self._e = e
self._log = (n - 1).bit_length()
self._size = 1 << self._log
self._d = [self._e()] * (self._size << 1)
if v is not None:
for i in range(self._n):
self._d[self._size + i] = v[i]
for i in range(self._size - 1, 0, -1):
self._d[i] = self._op(self._d[i << 1], self._d[i << 1 | 1])
def set(self, p, x):
p += self._size
self._d[p] = x
while p:
l, r = p, p^1
if l > r: l, r = r, l
self._d[p >> 1] = self._op(self._d[l], self._d[r])
p >>= 1
def get(self, p):
return self._d[p + self._size]
#[l, r)
def prod(self, l, r):
sml, smr = self._e(), self._e()
l += self._size
r += self._size
while l < r:
if l & 1:
sml = self._op(sml, self._d[l])
l += 1
if r & 1:
r -= 1
smr = self._op(self._d[r], smr)
l >>= 1
r >>= 1
return self._op(sml, smr)
def all_prod(self):
return self._d[1]
def op(x, y):
return x+y
def e():
return 0
from collections import deque
import sys
input = sys.stdin.readline
T = int(input())
for _ in range(T):
N = int(input())
P = list(map(int, input().split()))
deq = deque()
ST = SegTree(op, e, N)
x = 0
for i, p in enumerate(P):
cnt = ST.prod(0, p)
small = cnt
large = i-cnt
if small<large:
deq.appendleft(p)
x+=small
elif small>large:
deq.append(p)
x+=large
else:
if len(deq)>0 and deq[0]>p:
deq.appendleft(p)
else:
deq.append(p)
x+=small
ST.set(p-1, 1)
print(x)
print(*deq)
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