看了大神们的博客，仿着写的，地址：点击进入
Description
| 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
Output
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
#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;
}
POJ 2823 Sliding Window(单调队列),布布扣,bubuko.com
原文：http://blog.csdn.net/u013582254/article/details/38017561