poj 2823 Sliding Window

一 原题

Sliding Window

Time Limit: 12000MS		Memory Limit: 65536K
Total Submissions: 68479		Accepted: 19426
Case Time Limit: 5000MS

Description

An array of size  n ≤ 10 ⁶ 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

二 分析

裸的滑动窗口，优先队列和双端队列都可以

三 代码

优先队列：

/*
ID: maxkibble
LANG: c++
PROB: poj 2823
*/

#include <iostream>
#include <queue>

using namespace std;

const int maxn = 1e6 + 5;
int n, k, a[maxn], mn[maxn], mx[maxn];

struct cmp1 {
	bool operator () (const int &x, const int &y) {
		return a[x] > a[y];
	}
};

struct cmp2 {
	bool operator () (const int &x, const int &y) {
		return a[x] < a[y];	
	}
};

priority_queue<int, vector<int>, cmp1> q1;
priority_queue<int, vector<int>, cmp2> q2;

int main() {
	ios::sync_with_stdio(false);
	cin >> n >> k;
	for (int i = 0; i < n; i++) cin >> a[i];
	for (int i = 0; i < k - 1; i++) {
		q1.push(i); q2.push(i);
	}
	for (int i = k - 1; i < n; i++) {
		q1.push(i); q2.push(i);
		while (i - q1.top() > k - 1) q1.pop();
		while (i - q2.top() > k - 1) q2.pop();
		mn[i - k + 1] = a[q1.top()]; mx[i - k + 1] = a[q2.top()];
	}
	for (int i = 0; i < n - k + 1; i++) cout << mn[i] << " \n"[i == n - k];
	for (int i = 0; i < n - k + 1; i++) cout << mx[i] << " \n"[i == n - k];
	return 0;
}

双端队列：

/*
ID: maxkibble
LANG: c++
PROB: poj 2823
*/

#include <iostream>
#include <queue>

using namespace std;

#define mp make_pair
#define fi first
#define se second

typedef pair<int, int> PII;
const int maxn = 1e6 + 5;
int n, k, a[maxn], mn[maxn], mx[maxn];

deque<PII> q1, q2;

int main() {
	ios::sync_with_stdio(false);
	cin >> n >> k;
	for (int i = 0; i < n; i++) cin >> a[i];
	for (int i = 0, j = 0; i < n - k + 1; i++) {
		while (j < i + k) {
			while (!q1.empty() && q1.back().fi > a[j])
				q1.pop_back();
			while (!q2.empty() && q2.back().fi < a[j])
				q2.pop_back();
			q1.push_back(mp(a[j], j));
			q2.push_back(mp(a[j], j));
			j++;
		}
		while (q1.front().se < i) q1.pop_front();
		while (q2.front().se < i) q2.pop_front();
		mn[i] = q1.front().fi; mx[i] = q2.front().fi;
	}
	for (int i = 0; i < n - k + 1; i++) cout << mn[i] << " \n"[i == n - k];
	for (int i = 0; i < n - k + 1; i++) cout << mx[i] << " \n"[i == n - k];
	return 0;
}