poj 2823

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

分析：
这个题是单调队列的入门题。
求最大值：建立一个单调递减队列，元素从左到右依次入队，入队之前必须从队列尾部开始删除那些比当前入队元素小或者相等的元素，直到遇到一个比当前入队元素大的元素，或者队列为空为止。若此时队列的大小超过窗口值，则从队头删除元素，直到队列大小小入窗口值为止。然后把当前元素插入队尾。
求最小值：建立一个单调递增队列，元素从左到右依次入队，入队之前必须从队列发问开始删除那些比当前入队元素大或者相等的元素，直到遇到一个比当前入队元素小的元素，或者队列为空为止。若此时队列的大小超过窗口值，则从队头删除元素，直到队列大小小入窗口值为止。然后把当前元素插入队尾。
设窗口大小为k，本题解法为建立两个单调队列，先从原数组中取前k个元素按上述要求入队，然后获取队头元素，再把下一个元素放入队中，再获取头元素，这样一直到最后一个元素为止。最后把答案从头到尾依次输出即可。

代码：

var ans:array[0..1000000,1..2] of longint;    a,w:array[0..1000000] of longint;    h,f,r,mid,m,n,i,v:longint;begin  readln(n,m);  for i:=1 to n do    read(a[i]);  a[0]:=-2147483647;  for i:=1 to n do  begin    if a[i]>a[w[h]] then    begin      h:=h+1;      w[h]:=i;    end    else begin      f:=1;      r:=h;      repeat        mid:=(f+r) div 2;        if (a[w[mid]]>=a[i]) and (a[w[mid-1]]<a[i]) then break;        if (a[w[mid]]>=a[i]) then r:=mid-1 else f:=mid+1;      until f>r;      h:=mid;      w[h]:=i;    end;    if i>=m then    begin      f:=1;      r:=h;      v:=0;      repeat        mid:=(f+r) div 2;        if w[mid]<i-m+1 then f:=mid+1;        if w[mid]>=i-m+1 then        begin          v:=mid;          r:=mid-1;        end;      until f>r;      ans[i-m+1,1]:=a[w[v]];    end;  end;  a[0]:=2147483647;  h:=0;  fillchar(w,sizeof(w),0);  for i:=1 to n do  begin    if a[i]<a[w[h]] then    begin      h:=h+1;      w[h]:=i;    end    else begin      f:=1;      r:=h;      repeat        mid:=(f+r) div 2;        if (a[w[mid]]<=a[i]) and (a[w[mid-1]]>a[i]) then break;        if (a[w[mid]]<=a[i]) then r:=mid-1 else f:=mid+1;      until f>r;      h:=mid;      w[h]:=i;    end;    if i>=m then    begin      f:=1;      r:=h;      v:=0;      repeat        mid:=(f+r) div 2;        if w[mid]<i-m+1 then f:=mid+1;        if w[mid]>=i-m+1 then        begin          v:=mid;          r:=mid-1;        end;      until f>r;      ans[i-m+1,2]:=a[w[v]];    end;  end;  for i:=1 to n-m+1 do    write(ans[i,1],' ');  writeln;  for i:=1 to n-m+1 do    write(ans[i,2],' ');  writeln;end.