241. 楼兰图腾

Powered by:NEFU AB-IN

Link

241. 楼兰图腾

• 题意

  简略版：求数组中某个数，左边比它大的、小的有多少个，右边比它大的、小的有多少个

• 思路

  n有$2e5$，所以不能暴力判断，采用两遍树状数组，一次正序、一次逆序
  由于这里原数组是n的排列，所以不用离散化

• 代码

  '''
  Author: NEFU AB-IN
  Date: 2023-03-26 10:56:30
  FilePath: \Acwing\241\241.py
  LastEditTime: 2023-03-26 11:19:07
  '''
  # import
  import sys, math
  from collections import Counter, deque
  from heapq import heappop, heappush
  from bisect import bisect_left, bisect_right
  
  # Final
  N = int(2e5 + 10)
  INF = int(2e9)
  
  # Define
  sys.setrecursionlimit(INF)
  read = lambda: map(int, input().split())
  
  # —————————————————————Division line ————————————————————————————————————————
  
  tr = [0] * N
  
  
  def lowbit(x):
      return x & -x
  
  
  def add(x, v):
      while x < N:
          tr[x] += v
          x += lowbit(x)
  
  
  def query(x):
      res = 0
      while x:
          res += tr[x]
          x -= lowbit(x)
      return res
  
  
  class Point:
      def __init__(self, ls, ll, rs, rl):
          self.ls = ls
          self.ll = ll
          self.rs = rs
          self.rl = rl
  
  
  ans = [Point(0, 0, 0, 0) for _ in range(N)]
  
  n, = read()
  y = list(read())
  y1 = y[::-1]
  
  for i in range(n):
      # 左边比它小的
      ls = query(y[i] - 1)
      # 左边比它大的
      ll = i - query(y[i])
      add(y[i], 1)
      ans[i].ls = ls
      ans[i].ll = ll
  
  tr = [0] * N
  for i in range(n):
      # 右边比它小的
      rs = query(y1[i] - 1)
      # 右边比它大的
      rl = i - query(y1[i])
      add(y1[i], 1)
      ans[n - i - 1].rs = rs
      ans[n - i - 1].rl = rl
  
  res1, res2 = 0, 0
  
  for i in range(n):
      res1 += ans[i].ll * ans[i].rl
      res2 += ans[i].ls * ans[i].rs
  print(res1, res2)