看了大神们的博客，仿着写的，地址：点击进入
Description
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.
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
Source
POJ Monthly--2006.04.28, Ikki
额，单调队列的基础题吧，看了半天，想了半天，对于我这个没有学过数据结构的人来说有点难写。
从队头到队尾，元素在我们所关注的指标下是递减的（严格递减，而不是非递增），比如查询如果每次问的是窗口内的最小值，那么队列中元素从左至右就应该递增，如果每次问的是窗口内的最大值，则应该递减，依此类推。这是为了保证每次查询只需要取队头元素。
#include <iostream>
#include<algorithm>
#include<cstdio>
using namespace std;
const int maxn=1000001;
int Q[maxn];
int P[maxn];
int a[maxn];
int n,k;
void minQ()
{
int i;
int head=1,tail=0;
for(i=0; i<k-1; i++)//先入队
{
while(head<=tail && Q[tail]>=a[i])
--tail;
Q[++tail]=a[i];
P[tail]=i;
}
for(; i<n; i++)//for执行时每次都要输出一个答案=-=
{
while(head<=tail && Q[tail]>=a[i])
--tail;
Q[++tail]=a[i];
P[tail]=i;
while(P[head]<i-k+1)//移动到合法位置
head++;
printf(i==n-1?"%d\n":"%d ",Q[head]);
}
}
void maxQ()
{
int i;
int head=1,tail=0;
for(i=0;i<k-1;i++)
{
while(head<=tail&&Q[tail]<=a[i])
--tail;
Q[++tail]=a[i];
P[tail]=i;
}
for(;i<n;i++)
{
while(head<=tail&&Q[tail]<=a[i])
--tail;
Q[++tail]=a[i];
P[tail]=i;
while(P[head]<i-k+1)
head++;
printf(i==n-1?"%d\n":"%d ",Q[head]);
}
}
int main()
{
while(scanf("%d%d",&n,&k)==2)
{
for(int i=0;i<n;i++)
scanf("%d",&a[i]);
minQ();
maxQ();
}
return 0;
}