結果

問題 No.2422 regisys?
ユーザー sepa38
提出日時 2023-07-15 15:02:17
言語 PyPy3
(7.3.15)
結果
RE  
実行時間 -
コード長 1,838 bytes
コンパイル時間 138 ms
コンパイル使用メモリ 81,876 KB
実行使用メモリ 154,628 KB
最終ジャッジ日時 2024-09-16 23:56:54
合計ジャッジ時間 20,877 ms
ジャッジサーバーID
（参考情報） 		judge4 / judge3
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 1
other AC * 16 WA * 12 RE * 33
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 heapq
n, m = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
ls = [list(map(int, input().split())) for _ in range(n)]

def solve(n, m, a, b, ls):
  a2 = [[a[i], i] for i in range(n)]
  b2 = [[b[i], i] for i in range(n)]
  a2.sort()
  b2.sort()
  ls.sort(key = lambda x: x[1])
  flg = [0] * n
  h1 = []
  h2 = []
  idxa = 0
  idxb = 0
  for t, c in ls:
    while idxa < n:
      if a2[idxa][0] > c:
        break
      idx = a2[idxa][1]
      heapq.heappush(h1, (-b[idx])*n+idx)
      idxa += 1
    while idxb < n:
      if b2[idxb][0] > c:
        break
      idx = b2[idxb][1]
      heapq.heappush(h2, (-a[idx])*n+idx)
      idxb += 1
    if t:
      while h2:
        tmp = heapq.heappop(h2)
        x, i = divmod(tmp, n)
        if flg[i] == 0:
          flg[i] = 1
          break
    else:
      while h1:
        tmp = heapq.heappop(h1)
        x, i = divmod(tmp, n)
        if flg[i] == 0:
          flg[i] = 1
          break
    # print(flg, t, c)
    # print(h1, h2)
  return n - sum(flg)
  # print(flg)


from itertools import permutations
def naive(n, m, a, b, ls):
  ans = n
  for p in permutations(range(m)):
    res = n
    for i in range(m):
      t, c = ls[i]
      if t:
        res -= b[p[i]] <= c
      else:
        res -= a[p[i]] <= c
    ans = min(ans, res)
  return ans


print(solve(n, m, a, b, ls))
# print(naive(n, m, a, b, ls))

exit()


import random
for _ in range(10):
  n, m = random.randint(3, 5), random.randint(3, 5)
  m = max(m, n)
  a = [random.randint(1, 10) for _ in range(n)]
  b = [random.randint(1, 10) for _ in range(n)]
  ls = [[random.randint(0, 1), random.randint(1, 10)] for _ in range(m)]
  if solve(n, m, a, b, ls) != naive(n, m, a, b, ls):
    print(n, m)
    print(*a)
    print(*b)
    for t, c in ls:
      print(t, c)
    break
0