Sliding Window
An array of size n ≤ 10 6 is given to you. There is a sliding window of size kwhich 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题意：求区间宽为M，从左到右平移，求每个区间的最大值，最小值。
解法单调队列：
单调队列顾名思义维护一个单调减或单调增的队列，为了得到这个区间的最大值最小值，我们只需要和队头去比较就行了。
要的得到最小值，就要维护单调增的，因为这样队头才最小。
然后这个题还要注意，每个元素必须要在 那一时刻的区间类，队列的每个元素的小下标必须维护大于区间前端。
接下来数组模拟就行啦……
#include <iostream>
#include <string.h>
#include <stdio.h>
#include <cmath>
using namespace std;
typedef long long ll;
const int INF=0x3f3f3f3f;
const int maxn=1e6+7;
int n,m;
struct node
{
int id;
int val;
}q[maxn];
int ansmin[maxn];
int ansmax[maxn];
int a[maxn];
int head,tail;
void getmin()
{
///构造一个单调增的队列保证队头的元素始终是最小的，每次只比较队头
///先将前k-1个入队
int i;
head=1;
tail=0;
for(i=1;i<m;++i)
{
while(head<=tail&&q[tail].val>=a[i])/// 队尾元素大于将要入队的元素
--tail;
q[++tail].val=a[i];
q[tail].id=i;
}
for(;i<=n;++i)
{
while(head<=tail&&q[tail].val>=a[i])/// 队尾元素大于将要入队的元素
--tail;
q[++tail].val=a[i];
q[tail].id=i;
while(q[head].id<i-m+1)
{
++head;
}
ansmin[i-m+1]=q[head].val;
}
for(int i=1;i<n-m+1;++i)
cout<<ansmin[i]<<' ';
cout<<ansmin[n-m+1]<<endl;
}
void getmax()
{
int i;
head=1;
tail=0;
///构造单调减的队列
for(i=1;i<m;++i)
{
while(head<=tail&&q[tail].val<=a[i])
tail--;
q[++tail].id=i;
q[tail].val=a[i];
}
for(;i<=n;++i)
{
while(head<=tail&&q[tail].val<=a[i])
tail--;
q[++tail].id=i;
q[tail].val=a[i];
while(q[head].id<i-m+1)
++head;
ansmax[i-m+1]=q[head].val;
}
for(int i=1;i<n-m+1;++i)
cout<<ansmax[i]<<' ';
cout<<ansmax[n-m+1]<<endl;
}
int main()
{
while(scanf("%d%d",&n,&m)!=EOF)
{
for(int i=1;i<=n;++i)
scanf("%d",&a[i]);
getmin();
getmax();
}
return 0;
}