結果

問題 No.2422 regisys?
ユーザー とりゐ
提出日時 2023-08-12 14:58:02
言語 PyPy3
(7.3.15)
結果
AC  
実行時間 1,096 ms / 2,000 ms
コード長 4,827 bytes
コンパイル時間 309 ms
コンパイル使用メモリ 82,376 KB
実行使用メモリ 142,352 KB
最終ジャッジ日時 2024-11-19 22:26:04
合計ジャッジ時間 24,742 ms
ジャッジサーバーID
(参考情報)
judge5 / judge2
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 * 61
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

from sys import stdin
input=lambda :stdin.readline()[:-1]
import math
from bisect import bisect_left, bisect_right, insort
from typing import Generic, Iterable, Iterator, TypeVar, Union, List
T = TypeVar('T')
class SortedMultiset(Generic[T]):
BUCKET_RATIO = 50
REBUILD_RATIO = 170
def _build(self, a=None) -> None:
"Evenly divide `a` into buckets."
if a is None: a = list(self)
size = self.size = len(a)
bucket_size = int(math.ceil(math.sqrt(size / self.BUCKET_RATIO)))
self.a = [a[size * i // bucket_size : size * (i + 1) // bucket_size] for i in range(bucket_size)]
def __init__(self, a: Iterable[T] = []) -> None:
"Make a new SortedMultiset from iterable. / O(N) if sorted / O(N log N)"
a = list(a)
if not all(a[i] <= a[i + 1] for i in range(len(a) - 1)):
a = sorted(a)
self._build(a)
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 __len__(self) -> int:
return self.size
def __repr__(self) -> str:
return "SortedMultiset" + str(self.a)
def __str__(self) -> str:
s = str(list(self))
return "{" + s[1 : len(s) - 1] + "}"
def _find_bucket(self, x: T) -> List[T]:
"Find the bucket which should contain x. self must not be empty."
for a in self.a:
if x <= a[-1]: return a
return a
def __contains__(self, x: T) -> bool:
if self.size == 0: return False
a = self._find_bucket(x)
i = bisect_left(a, x)
return i != len(a) and a[i] == x
def 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 = 1
return
a = self._find_bucket(x)
insort(a, x)
self.size += 1
if len(a) > len(self.a) * self.REBUILD_RATIO:
self._build()
def discard(self, x: T) -> bool:
"Remove an element and return True if removed. / O(√N)"
if self.size == 0: return False
a = self._find_bucket(x)
i = bisect_left(a, x)
if i == len(a) or a[i] != x: return False
a.pop(i)
self.size -= 1
if len(a) == 0: self._build()
return True
def lt(self, x: T) -> Union[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) -> Union[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) -> Union[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) -> Union[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, x: int) -> T:
"Return the x-th element, or IndexError if it doesn't exist."
if x < 0: x += self.size
if x < 0: raise IndexError
for a in self.a:
if x < len(a): return a[x]
x -= 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
n,m=map(int,input().split())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
t0=[]
t1=[]
for i in range(m):
t,c=map(int,input().split())
if t:
t1.append(c)
else:
t0.append(c)
t0.sort()
t1.sort()
from heapq import heappop, heappush
ab=[]
for i in range(n):
ab.append([a[i],b[i]])
ab.sort()
idx=0
hq=[]
ans=0
for c in t0:
while idx!=n and ab[idx][0]<=c:
heappush(hq,-ab[idx][1])
idx+=1
if hq:
ans+=1
heappop(hq)
for i in range(idx,n):
heappush(hq,-ab[i][1])
hq2=[]
for i in hq:
heappush(hq2,-i)
for c in t1:
if hq2 and hq2[0]<=c:
heappop(hq2)
ans+=1
print(n-ans)
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