POJ 2823 Sliding Window（单调队列）

单调队列
单调队列就是一个其中的数具有单调性的数列，只能从队尾入列，而可以从队头和队尾出列。
这里的单调性是指从队头到队尾数值的变化情况。
求最大值是建立一个单调递减队列，这样队头就是最大的元素，当数量超过规定时，队头出队。
求最小值是建立一个单调递增队列，这样队头就是最小的元素，当数量超过规定时，队头出队。

POJ 2823 Sliding Window
An array of size n ≤ 10 6 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.
Input
The input consists of two lines. The first line contains two integers n and k which are the lengths of the array and the sliding window. There are n integers in the second line.
Output
There are two lines in the output. The first line gives the minimum values in the window at each position, from left to right, respectively. The second line gives the maximum values.
Sample Input
8 3
1 3 -1 -3 5 3 6 7
Sample Output
-1 -3 -3 -3 3 3
3 3 5 5 6 7

#include <iostream>
#include <cstdio>
#include <queue>
#include <cstring>
#include <algorithm>
using namespace std;

int a[1000005], q1[1000005], q2[1000005];
int n, m, head, tail;
int main()
{
  scanf("%d%d", &n, &m);
  for (int i=1; i<=n; i++)
    scanf("%d", &a[i]);
  head=1, tail=0;
  for (int i=1; i<=n; i++){
    while (head<=tail && a[q1[tail]]>a[i])
      tail--;
    while (head<=tail && i-q1[head]+1>m)  
      head++;
    q1[++tail]=i;
    if (i>=m) printf("%d ", a[q1[head]]);
  }
  printf("\n");
  
  head=1, tail=0;
  for (int i=1; i<=n; i++){
    while (head<=tail && a[q2[tail]]<a[i])
      tail--;
    while (head<=tail && i-q2[head]+1>m)
      head++;
    q2[++tail]=i;
    if (i>=m) printf("%d ", a[q2[head]]); 
  }
  return 0;
}