POJ 2823 Sliding Window ST RMQ

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

Source

POJ monthly--2006.04.28, Ikki Test Instructions:given a sequence of long n, ask the minimum and maximum of all the cases in the series with a length of K. Ideas:Seniors Teach the RMQ solution, St version of the essence is DP, than do not understand DP before, now feel good understanding more. In addition, you can use segment tree solutions. Note the Log table first.

1#include <stdio.h>2#include <algorithm>3 //#define LOG[I] = (I & (i-1))? LOG[I-1]: log[i-1] + 14 #defineMAXX 12345675#include <vector>6 using namespacestd;7 8 intA[maxx];9 intdp1[maxx][ A];Ten intLog[maxx]; One  A voidInitintN) - { -log[1] =0;  the      for(intI=2; i<=n; i++) -log[i]= (i& (i-1))? log[i-1]:log[i-1]+1; - } -  + intST (intLintRinti) - { +     intk=log[r-l+1]; A     if(i==1) at         returnMax (dp1[l][k],dp1[r-(1&LT;&LT;K) +1][k]); -     if(i==0) -         returnMin (dp1[l][k],dp1[r-(1&LT;&LT;K) +1][k]); - } - intMain () - { in     intN, K; -      to      while(~SCANF ("%d%d", &n, &k)) +     { -         intI, J; the init (n); *          for(i=1; i<=n; i++) $         {Panax Notoginsengscanf"%d", &a[i]); -dp1[i][0]=A[i]; the         } +          for(j=1; j<= -; J + +) A         {  the              for(i=1; i<=n; i++) +             { -                 if(I+ (1&LT;&LT;J)-1>N) $                      Break; $Dp1[i][j]=min (dp1[i][j-1], dp1[i+ (1<< (J-1))][j-1]); -             } -         } the          for(i=1; i<=n-k+1; i++) -         {Wuyi             if(i!=1) theprintf" "); -printf"%d", ST (i,i+k-1,0)); Wu         } -         ////// About  $          for(i=1; i<=n; i++) -         { -dp1[i][0]=A[i]; -              for(j=1; j<= -; J + +) Adp1[i][j]=0; +         } the          for(j=1; j<= -; J + +) -         {  $              for(i=1; i<=n; i++) the             { the                 if(I+ (1&LT;&LT;J)-1>N) the                      Break; theDp1[i][j]=max (dp1[i][j-1], dp1[i+ (1<< (J-1))][j-1]); -             } in         }  theprintf"\ n"); the          for(i=1; i<=n-k+1; i++) About         { the             if(i!=1) theprintf" "); theprintf"%d", ST (i,i+k-1,1)); +         } -printf"\ n"); the     }Bayi} 

POJ 2823 Sliding Window ST RMQ