Poj 2823 单调队列 Sliding Window

最新推荐文章于 2022-10-22 22:10:29 发布

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最新推荐文章于 2022-10-22 22:10:29 发布

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分类专栏：单调队列

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本文介绍了一种解决滑动窗口最大值与最小值问题的方法，通过使用单调队列进行高效计算。对于长度为 n 的数组及大小为 k 的滑动窗口，每次移动窗口时都能快速找出窗口内的最大值与最小值。

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An array of size n ≤ 10⁶ is given to you. There is a sliding window of size k which is moving from the very left of the array to the very right. You can only see the k numbers in the window. Each time the sliding window moves rightwards by one position. Following is an example:
The array is [1 3 -1 -3 5 3 6 7], and k is 3.

Window position	Minimum value	Maximum value
[1 3 -1] -3 5 3 6 7	-1	3
1 [3 -1 -3] 5 3 6 7	-3	3
1 3 [-1 -3 5] 3 6 7	-3	5
1 3 -1 [-3 5 3] 6 7	-3	5
1 3 -1 -3 [5 3 6] 7	3	6
1 3 -1 -3 5 [3 6 7]	3	7

Your task is to determine the maximum and minimum values in the sliding window at each position.

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朴素的单调队列。维护一个单调队列，使其队首元素在窗口内（i-q[head]>=k时出队），且保证单调不上升。对于min只要取相反数即可。此处有坑。。。k=1。。。。。。

var
  n,k,i,head,tail:longint;
  q,a:array[1..1000000] of longint;
begin
  read(n,k);
  if (n=0) or (k>n)  then exit;
  for i:=1 to n do read(a[i]);
  q[1]:=1;head:=1;tail:=1;
  if k=1 then write(a[1]);//之前被忽略。。。
  for i:=1 to n do a[i]:=-a[i];
  for i:=2 to n do begin
    if i-q[head]>=k then inc(head);
	while (tail>=head) and (a[q[tail]]<a[i]) do dec(tail);
	inc(tail);
	q[tail]:=i;
	if i=k then write(-a[q[head]]) else if i>k then write(' ',-a[q[head]]);
  end;           writeln;

  q[1]:=1;head:=1;tail:=1;
  for i:=1 to n do a[i]:=-a[i];
  if k=1 then write(a[1]);
  for i:=2 to n do begin
    if i-q[head]>=k then inc(head);
	while (tail>=head) and (a[q[tail]]<a[i]) do dec(tail);
	inc(tail);
	q[tail]:=i;
	if i=k then write(a[q[head]]) else if i>k then write(' ',a[q[head]]);
  end;           writeln;
end.