POJ 2823 Sliding window (the most-valued problem of sliding windows)

Sliding Window

Time Limit: 12000MS		Memory Limit: 65536K
Total Submissions: 41264		Accepted: 12229
Case Time Limit: 5000MS

Description

An array of size N≤106 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. can only see the kNumbers in the window. Each of the sliding window moves rightwards by one position. Following is an example:
The array is[1 3-1-3 5 3 6 7], and kis 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 the lines. The first line contains integers Nand kWhich is the lengths of the array and the sliding window. There is NIntegers in the second line.

Output

There is lines in the output. The first line gives the minimum values in the windows at each position, from left to right, respectively. The second line gives the maximum values.

Sample Input

8 31 3-1-3 5 3 6 7

Sample Output

-1-3-3-3 3 33 3 5 5 6 7
I did some optimization simulations according to my own thinking, and still timed out.
Code:

#include <stdio.h> #include <string.h> #include <stdlib.h> #include <algorithm> #include < Deque>using namespace Std;struct node{int mi;int ma;}  Q[1000004];int A[1000004];int Main () {int n, k;int i, j;scanf ("%d%d", &n, &k), for (int i=0; i<n; i++) scanf ("%d", &a[i]); int dd=a[0], ff=a[0];for (i=1; i<k; i++) {if (A[I]&GT;DD) dd=a[i];//maxif (A[I]&LT;FF) ff=a[i];//min}int e= 0;Q[E].MA=DD; Q[e++].mi=ff;int pos;for (j=k; j<n; j + +)//traverse these sequences {if (A[J]&GT;DD) dd=a[j];if (A[J]&LT;FF) ff=a[j];//Determine whether to update pos=j-k; if (a[pos]>dd && a[pos]<ff) Continue;else{int p, w;if (A[POS]==DD)//start is = = Max Update max value {p=a[pos+1];for (i=pos+ 2; i<=j; i++) {P=max (P, a[i]);} Dd=p;} else if (A[POS]==FF)//start is = = Minimum update minimum value {w=a[pos+1];for (i=pos+2; i<=j; i++) {w=min (W, a[i]);} Ff=w;} Q[E].MA=DD; Q[E++].MI=FF;}}        for (i=0; i<e; i++) {if (i==e-1) printf ("%d\n", Q[I].MI); else printf ("%d", Q[I].MI);} for (i=0; i<e; i++) {if (i==e-1) printf ("%d\n", Q[I].ma]; else printf ("%d", q[i].ma);} return 0;}

POJ 2823 Sliding window (the most-valued problem of sliding windows)