思路分析:

    設輸入序列為, 從左向右依次掃描輸入序列, 並維持兩個佇列minde和maxde, 當考察時, 其中佇列minde和maxde中至多含有k個元素, minde中元素嚴格單調遞增, maxde中元素嚴格單調遞減, 具體如下AC程式碼, 由於輸入序列中每個元素至多入佇列一次和出佇列一次, 因此下面程式的時間複雜度為O(n)

//POJ2823_Sliding Window
#include <iostream>
#include <cstdio>
#include <deque>
#include <algorithm>
using namespace std;
const int MAX = 1e6 + 5;
int arr[MAX], n, k, minval[MAX], maxval[MAX];
deque<int> minde, maxde;
int main(){
	scanf("%d %d", &n, &k);	
	for(int i = 1; i <= n; ++i) scanf("%d", &arr[i]);
	minval[1] = maxval[1] = arr[1], minde.push_back(1), maxde.push_back(1);
	for(int i = 2; i <= n; ++i){
		if(minde[0] < i - k + 1) minde.pop_front();
		if(maxde[0] < i - k + 1) maxde.pop_front();
		while(!minde.empty() && arr[minde.back()] >= arr[i]) minde.pop_back();
		minde.push_back(i), minval[i] = arr[minde[0]];
		while(!maxde.empty() && arr[maxde.back()] <= arr[i]) maxde.pop_back();
		maxde.push_back(i), maxval[i] = arr[maxde[0]];
	}
	for(int i = k; i <= n; ++i) cout << minval[i] << " "; cout << endl;
	for(int i = k; i <= n; ++i) cout << maxval[i] << " "; cout << endl;
	return 0;
}