POJ 2823 Sliding Window ST RMQ

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 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

Source

POJ Monthly--2006.04.28, Ikki

 

题意 ：给定长n的数列，问长为k的区间在数列中所有情况的最小值和最大值。

思路：学长教导的RMQ解法，ST版实质是DP，比起不太懂DP的以前，现在感觉好理解多了。此外感觉可以使用线段树解。

注意先打log的表。

 #include <stdio.h>
 #include <algorithm>
 //#define LOG[i] = (i & (i - 1)) ? LOG[i - 1] : LOG[i - 1] + 1
 #define MAXX 1234567
 #include <vector>
 using namespace std;

 int a[MAXX];
 int dp1[MAXX][];
 int LOG[MAXX];

 void init(int n)
 {
     LOG[] = ;
     for(int i=; i<=n; i++)
         LOG[i]=(i&(i-))?LOG[i-]:LOG[i-]+;
 }

 int ST(int l, int r, int i)
 {
     int k=LOG[r-l+];
     if(i==)
         return max(dp1[l][k],dp1[r-(<<k)+][k]);
     if(i==)
         return min(dp1[l][k],dp1[r-(<<k)+][k]);
 }
 int main()
 {
     int n, k;

     while(~scanf("%d%d",&n, &k))
     {
         int i, j;
         init(n);
         for(i=; i<=n; i++)
         {
             scanf("%d", &a[i]);
             dp1[i][]=a[i];
         }
         for(j=; j<=; j++)
         {
             for(i=; i<=n; i++)
             {
                 if(i+(<<j)->n)
                     break;
                 dp1[i][j]=min(dp1[i][j-], dp1[i+(<<(j-))][j-]);
             }
         }
         for(i=; i<=n-k+; i++)
         {
             if(i!=)
                 printf(" ");
             printf("%d", ST(i,i+k-,));
         }
         //////

         for(i=; i<=n; i++)
         {
             dp1[i][]=a[i];
             for(j=; j<=; j++)
                 dp1[i][j]=;
         }
         for(j=; j<=; j++)
         {
             for(i=; i<=n; i++)
             {
                 if(i+(<<j)->n)
                     break;
                 dp1[i][j]=max(dp1[i][j-], dp1[i+(<<(j-))][j-]);
             }
         }
         printf("\n");
         for(i=; i<=n-k+; i++)
         {
             if(i!=)
                 printf(" ");
             printf("%d", ST(i,i+k-,));
         }
         printf("\n");
     }
 }