﻿ Sliding Window（滑动窗口）

### Sliding Window（滑动窗口）

 Time Limit: 12000MS Memory Limit: 65536K Total Submissions: 58002 Accepted: 16616 Case Time Limit: 5000MS

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.
Window positionMinimum valueMaximum 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

## 题目描述

The array is [1 3 -1 -3 5 3 6 7], and k = 3.

## 输入输出样例

```8 3
1 3 -1 -3 5 3 6 7```

```-1 -3 -3 -3 3 3
3 3 5 5 6 7```

50%的数据，n<=10^5

100%的数据，n<=10^6

## 代码实现：

``` 1 #include<cstdio>
2 #define maxn 1000010
3 int n,k,s[maxn];
4 int a;
5 int h1,t1,q1[maxn],a1[maxn],l1;
6 int h2,t2,q2[maxn],a2[maxn],l2;
7 int main(){
8     scanf("%d%d",&n,&k);
9     for(int i=1;i<=n;i++){
10         scanf("%d",&s[i]);
11         a=h1-1;h1=t1;
12         for(int j=a;j>=t1;j--) if(s[i]>=q1[j]){h1=j+1;break;}
13         q1[h1++]=s[i];
14         if(i>k&&s[i-k]==q1[t1]) t1++;//因为判断先后的问题，一直RE一个点。
15         if(i>=k) a1[l1++]=q1[t1];
16         a=h2-1;h2=t2;
17         for(int j=a;j>=t2;j--) if(s[i]<=q2[j]){h2=j+1;break;}
18         q2[h2++]=s[i];
19         if(i>k&&s[i-k]==q2[t2]) t2++;
20         if(i>=k) a2[l2++]=q2[t2];
21     }
22     for(int i=0;i<l1;i++) printf("%d ",a1[i]);putchar('
');
23     for(int i=0;i<l2;i++) printf("%d ",a2[i]);putchar('
');
24     return 0;
25 }```

``` 1 #include<cstdio>
2 #define maxn 1000010
3 int n,k,s[maxn];
4 int a;
5 int h1,t1,q1[maxn],a1[maxn],l1;
6 int h2,t2,q2[maxn],a2[maxn],l2;
7 int main(){
8     scanf("%d%d",&n,&k);
9     for(int i=1;i<=n;i++){
10         scanf("%d",&s[i]);
11         q1[h1++]=1000000000;q2[h2++]=-1000000000;
12         for(int j=t1;j<h1;j++)
13         if(s[i]<q1[j]){q1[j]=s[i];h1=j+1;break;}
14         if(i>k&&s[i-k]==q1[t1]) t1++;
15         if(i>=k) a1[l1++]=q1[t1];
16         for(int j=t2;j<h2;j++)
17         if(s[i]>q2[j]){q2[j]=s[i];h2=j+1;break;}
18         if(i>k&&s[i-k]==q2[t2]) t2++;
19         if(i>=k) a2[l2++]=q2[t2];
20     }
21     for(int i=0;i<l1;i++) printf("%d ",a1[i]);putchar('
');
22     for(int i=0;i<l2;i++) printf("%d ",a2[i]);putchar('
');
23     return 0;
24 }```

``` 1 #include<cstdio>
2 #define maxn 1000010
3 int n,k,a,h,t;
4 int s[maxn],q[maxn];
5 int main(){
6     scanf("%d%d",&n,&k);
7     for(int i=1;i<=n;i++){
8         scanf("%d",&s[i]);
9         q[h++]=1000000000;
10         for(int j=t;j<h;j++) if(s[i]<q[j]){q[j]=s[i];h=j+1;break;}
11         if(i>k&&s[i-k]==q[t]) t++;
12         if(i>=k) printf("%d ",q[t]);
13     }
14     putchar('
');h=t=0;
15     for(int i=1;i<=n;i++){
16         q[h++]=-1000000000;
17         for(int j=t;j<h;j++) if(s[i]>q[j]){q[j]=s[i];h=j+1;break;}
18         if(i>k&&s[i-k]==q[t]) t++;
19         if(i>=k) printf("%d ",q[t]);
20     }
21     putchar('
');
22     return 0;
23 }```

## 最后被poj G++接受的代码（2266mm）：

``` 1 #include<cstdio>
2 #define maxn 1000010
3 int n,k,s[maxn];
4 int a;
5 int h1,t1,q1[maxn],a1[maxn],l1;
6 int h2,t2,q2[maxn],a2[maxn],l2;
7 inline int abs(int x){return x<0?-x:x;}
8 void write(int x){
9     if(x) write(x/10);
10     else return;
11     putchar(x%10+'0');
12 }
13 int main(){
14     scanf("%d%d",&n,&k);
15     for(int i=1;i<=n;i++){
16         scanf("%d",&s[i]);
17         a=h1-1;h1=t1;
18         for(int j=a;j>=t1;j--) if(s[i]>=q1[j]){h1=j+1;break;}
19         q1[h1++]=s[i];
20         if(i>k&&s[i-k]==q1[t1]) t1++;
21         if(i>=k) a1[l1++]=q1[t1];
22         a=h2-1;h2=t2;
23         for(int j=a;j>=t2;j--) if(s[i]<=q2[j]){h2=j+1;break;}
24         q2[h2++]=s[i];
25         if(i>k&&s[i-k]==q2[t2]) t2++;
26         if(i>=k) a2[l2++]=q2[t2];
27     }
28     for(int i=0;i<l1;i++){
29         if(a1[i]<0) putchar('-');
30         write(abs(a1[i]));
31         if(!a1[i]) putchar('0');
32         if(i<l1-1) putchar(' ');
33     }
34     putchar('
');
35     for(int i=0;i<l2;i++){
36         if(a2[i]<0) putchar('-');
37         write(abs(a2[i]));
38         if(!a2[i]) putchar('0');
39         if(i<l2-1) putchar(' ');
40     }
41     putchar('
');
42     return 0;
43 }```

``` 1 #include<cstdio>
2 const int maxn=4e6;
3 int n,k;
4 int s[maxn];
5 int t_min[maxn],t_max[maxn];
6 inline int min_(int x,int y){return x<y?x:y;}
7 inline int max_(int x,int y){return x>y?x:y;}
8 void build_min(int k,int l,int r){
9     if(l==r){
10         t_min[k]=s[l];
11         return;
12     }
13     int mid=l+r>>1,ls=k<<1,rs=ls|1;
14     build_min(ls,l,mid);
15     build_min(rs,mid+1,r);
16     t_min[k]=t_min[ls]<t_min[rs]?t_min[ls]:t_min[rs];
17 }
18 int search_min(int k,int l,int r,int al,int ar){
19     if(al==l&&ar==r) return t_min[k];
20     int ret=1e9,mid=l+r>>1,ls=k<<1,rs=ls|1;
21     if(al<=mid) ret=min_(ret,search_min(ls,l,mid,al,min_(ar,mid)));
22     if(ar>mid) ret=min_(ret,search_min(rs,mid+1,r,max_(al,mid+1),ar));
23     return ret;
24 }
25 void build_max(int k,int l,int r){
26     if(l==r){
27         t_max[k]=s[l];
28         return;
29     }
30     int mid=l+r>>1,ls=k<<1,rs=ls|1;
31     build_max(ls,l,mid);
32     build_max(rs,mid+1,r);
33     t_max[k]=t_max[ls]>t_max[rs]?t_max[ls]:t_max[rs];
34 }
35 int search_max(int k,int l,int r,int al,int ar){
36     if(al==l&&ar==r) return t_max[k];
37     int ret=-1e9,mid=l+r>>1,ls=k<<1,rs=ls|1;
38     if(al<=mid) ret=max_(ret,search_max(ls,l,mid,al,min_(ar,mid)));
39     if(ar>mid) ret=max_(ret,search_max(rs,mid+1,r,max_(al,mid+1),ar));
40     return ret;
41 }
42 int main(){
43     scanf("%d%d",&n,&k);
44     for(int i=1;i<=n;i++) scanf("%d",&s[i]);
45     build_min(1,1,n);
46     build_max(1,1,n);
47     for(int i=1;i<=n-k+1;i++)
48     printf("%d ",search_min(1,1,n,i,i+k-1));
49     putchar('
');
50     for(int i=1;i<=n-k+1;i++)
51     printf("%d ",search_max(1,1,n,i,i+k-1));
52     putchar('
');
53     return 0;
54 }```

## 常数优化十分可观的zkw线段树(洛谷 2275ms)

``` 1 #include<cstdio>
2 const int maxn=1<<22;
3 int n,k,m;
4 int t[maxn];
5 inline int min_(int x,int y){return x<y?x:y;}
6 inline int max_(int x,int y){return x>y?x:y;}
7 void build_min(){for(int i=m-1;i>0;i--) t[i]=min_(t[i<<1],t[i<<1|1]);}
8 int search_min(int l,int r,int ret){
9     for(l+=m,r+=m;r-l!=1;l>>=1,r>>=1){
10         if(~l&1) ret=min_(ret,t[l^1]);
11         if(r&1) ret=min_(ret,t[r^1]);
12     }
13     return ret;
14 }
15 void build_max(){for(int i=m-1;i>0;i--) t[i]=max_(t[i<<1],t[i<<1|1]);}
16 int search_max(int l,int r,int ret){
17     for(l+=m,r+=m;r-l!=1;l>>=1,r>>=1){
18         if(~l&1) ret=max_(ret,t[l^1]);
19         if(r&1) ret=max_(ret,t[r^1]);
20     }
21     return ret;
22 }
23 int main(){
24     scanf("%d%d",&n,&k);
25     for(m=1;m<=n+1;m<<=1);
26     for(int i=1;i<=n;i++) scanf("%d",t+m+i);
27     build_min();
28     for(int i=0;i<=n-k;i++)
29     printf("%d ",search_min(i,i+k+1,1e9));
30     putchar('
');
31     build_max();
32     for(int i=0;i<=n-k;i++)
33     printf("%d ",search_max(i,i+k+1,-1e9));
34     putchar('
');
35     return 0;
36 }```