poj-2823 Sliding Window(单调队列)

最新推荐文章于 2021-01-25 20:02:52 发布

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本文介绍了一种高效算法，用于求解给定数组中滑动窗口的最大值与最小值问题。通过使用两个模拟队列，分别维护窗口内的最大值和最小值，实现了O(n)的时间复杂度。

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原题链接

Sliding Window

Time Limit: 12000MS		Memory Limit: 65536K
Total Submissions: 54929		Accepted: 15814
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

用数组模拟单调递减队列deq1，单调递增队列deq2，deq1中的首元素是最大值，deq2中的首元素是最小值，再取队首元素时判断对首元素是否在k个元素的区间内

如果不是则队首指针+1

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <stack>
#include <deque> 
#define INF 1e9
#define maxn 1000005
using namespace std;
typedef long long ll;

int num[maxn], maxs[maxn], mins[maxn];
int deq1[maxn], deq2[maxn];
int main(){
	
//	freopen("in.txt", "r", stdin);
	int n, k;
	while(scanf("%d%d", &n, &k) == 2){
		int l1 = 0, r1 = 0, l2 = 0, r2 = 0; 
		if(k <= 0)
		 continue;
		for(int i = 0; i < n; i++)
		 scanf("%d", num+i);
		for(int i = 0; i < n; i++){
			while(l1 != r1 && num[deq1[r1-1]] <= num[i])
			 r1--;
			deq1[r1++] = i;
		    while(l2 != r2 && num[deq2[r2-1]] >= num[i])
			 r2--;
			deq2[r2++] = i;
			if(i >= k-1){
				int j = i - k;
				while(deq1[l1] <= j)
				 l1++;
				while(deq2[l2] <= j)
				 l2++;
				maxs[i] = num[deq1[l1]];
				mins[i] = num[deq2[l2]];
			}
		}
		printf("%d", mins[k-1]);
		for(int i = k; i < n; i++)
		 printf(" %d", mins[i]);
		puts("");
		printf("%d", maxs[k-1]);
		for(int i = k; i < n; i++)
		 printf(" %d", maxs[i]);
		puts("");
	}
	return 0;
}