poj 2823 单调队列 deque写法_deque结构体写法-优快云博客

本文详细介绍了滑动窗口算法的基本概念、实现细节及在实际问题中的优化策略。通过实例演示了如何高效地计算数组中子数组的最大值和最小值，提供了多种优化技巧以提升算法性能。

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Sliding Window

Time Limit: 12000MS		Memory Limit: 65536K
Total Submissions: 46435		Accepted: 13417
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

#include <iostream>
#include <cstdio>
#include <queue>
#include <deque>

using namespace std;
typedef pair<int, int> P;
#define maxn 1000000 + 10

deque<P> Q1;
deque<P> Q2;
int n, k;
int Min[maxn], Max[maxn];

int main()
{
    while(~scanf("%d%d", &n, &k))
    {
        while(!Q1.empty()) Q1.pop_back();
        while(!Q2.empty()) Q2.pop_back();
        int x;
        for(int i=1; i<=n; i++)
        {
            scanf("%d", &x);
            while(!Q1.empty() && Q1.back().first >= x) Q1.pop_back();
            Q1.push_back(P(x, i));
            if(i >= k)
                {
                    while(!Q1.empty() && Q1.front().second <= i-k) Q1.pop_front();
                    Min[i] = Q1.front().first;
                }

            while(!Q2.empty() && Q2.back().first <= x) Q2.pop_back();
            Q2.push_back(P(x, i));
            if(i >= k)
                {
                    while(!Q2.empty() && Q2.front().second <= i-k) Q2.pop_front();
                    Max[i] = Q2.front().first;
                }
        }

        for(int i=k; i<=n; i++)
            i == n ? printf("%d\n", Min[i]) : printf("%d ", Min[i]);

        for(int i=k; i<=n; i++)
            i == n ? printf("%d\n", Max[i]) : printf("%d ", Max[i]);
    }
    return 0;
}