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

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 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. 

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. 

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. 

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

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

POJ Monthly--2006.04.28, Ikki

 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(‘\n‘);
23     for(int i=0;i<l2;i++) printf("%d ",a2[i]);putchar(‘\n‘);
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(‘\n‘);
22     for(int i=0;i<l2;i++) printf("%d ",a2[i]);putchar(‘\n‘);
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(‘\n‘);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(‘\n‘);
22     return 0;
23 }

 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(‘\n‘);
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(‘\n‘);
42     return 0;
43 }