Модуль: حساب التعقيد المقارب

 # include & lt؛ bits / stdc ++. h & gt؛ 
 باستخدام   مساحة الاسم  std؛

 int   الرئيسي  ()
{
 int  n، m؛
المتجه  & lt؛  ناقل  & lt؛   int   & gt؛   & gt؛  up1 (n ، vector  & lt؛   int   & gt؛  (m))؛
 int  ans  =   0 ؛
 لـ  ( int  i  =   1 ؛ i  & lt؛ =  n؛ i  ++ )
{
المتجه  & lt؛   int   & gt؛  L (m  +   1 ،  1 < / span>)، R (m  +   1 ، m)؛
مكدس  & lt؛   int   & gt؛  ف ؛
 لـ  ( int  j  =   1 ؛ j  & lt؛ =  m؛ j  ++ )
{
 while  (!  q.empty ()  & amp؛ & amp؛  up1 [i] [j]  & lt؛  up1 [i] [q.top ()])
{
R [q.top ()]  =  j  -   1  ؛
q.pop () ؛
}
qpush (ي) ؛
}
 بينما  (!  q.empty ())
q.pop () ؛
 لـ  ( int  j  =  m؛ j  & gt؛ =   1 ؛ j  - )
{
 while  (!  q.empty ()  & amp؛ & amp؛  up1 [i] [j]  & lt؛  up1 [i] [q.top ()])
{
L [q.top ()]  =  j  +   1  ؛
q.pop () ؛
}
qpush (ي) ؛
}
 لـ  ( int  j  =   1 ؛ j  & lt؛ =  m؛ j  ++ )
ans  =  max (ans، up1 [i] [j]  *  (R [j] < span style = "color: # 666666"> -  L [j]  +   1 )) ؛
}
cout  & lt؛ & lt؛  ans؛
 return   0 ؛
}