Poj 2823 单调队列 Sliding Window

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. 
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朴素的单调队列。维护一个单调队列，使其队首元素在窗口内（i-q[head]>=k时出队），且保证单调不上升。对于min只要取相反数即可。                                                                                 此处有坑。。。k=1。。。。。。

var
n,k,i,head,tail:longint;
q,a:array[1..1000000] of longint;
begin
read(n,k);
if (n=0) or (k>n)  then exit;
for i:=1 to n do read(a[i]);
q[1]:=1;head:=1;tail:=1;
if k=1 then write(a[1]);//之前被忽略。。。
for i:=1 to n do a[i]:=-a[i];
for i:=2 to n do begin
if i-q[head]>=k then inc(head);
while (tail>=head) and (a[q[tail]]<a[i]) do dec(tail);
inc(tail);
q[tail]:=i;
if i=k then write(-a[q[head]]) else if i>k then write(' ',-a[q[head]]);
end;           writeln;

q[1]:=1;head:=1;tail:=1;
for i:=1 to n do a[i]:=-a[i];
if k=1 then write(a[1]);
for i:=2 to n do begin
if i-q[head]>=k then inc(head);
while (tail>=head) and (a[q[tail]]<a[i]) do dec(tail);
inc(tail);
q[tail]:=i;
if i=k then write(a[q[head]]) else if i>k then write(' ',a[q[head]]);
end;           writeln;
end.