POJ2823Sliding Window(单调队列）

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本文介绍了一种解决滑动窗口最值问题的有效算法——单调队列。通过维护两个单调队列，分别找出滑动窗口内的最大值和最小值。适用于处理大量数据中的滑动窗口计算任务。

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Sliding Window

Time Limit: 12000MS Memory Limit: 65536K
Total Submissions: 55878 Accepted: 16068
Case Time Limit: 5000MS

Description

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.

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

题意：有n个数，和一个长度为k的窗户。每次得到k个数的最值。
很裸的单调队列。在代码中给出具体解释。
代码：

#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include <queue>
#include <stack>
#include <map>
#include <algorithm>
#include <vector>
#define bababaa printf("!!!!!!!\n")
using namespace std;
const int  N=1e6+10;
int num[N],mx[N],mi[N];
int main()
{
    int n,k,len1,len2;
    len1=len2=0;
    deque<int>q1,q2;
    scanf("%d%d",&n,&k);
    for(int i=1;i<=n;i++)
        scanf("%d",&num[i]);
    for(int i=1;i<=n;i++)
    {
        if(q1.empty()) q1.push_back(i);
        else
        {
            while(!q1.empty()&&(num[q1.back()]<num[i])) q1.pop_back();//求最大值要维护一单调递减队列。
            while(!q1.empty()&&i-q1.front()+1>k) q1.pop_front();
            q1.push_back(i);
        }
        if(i>=k)
            mx[len1++]=num[q1.front()];
    }
    for(int j=1;j<=n;j++)
    {
         if(q2.empty()) q2.push_back(j);
           else
           {
               while(!q2.empty()&&(num[q2.back()]>num[j])) q2.pop_back();
               while(!q2.empty()&&j-q2.front()+1>k) q2.pop_front();
               q2.push_back(j);
            }
            if(j>=k) mi[len2++]=num[q2.front()];

    }
   for(int i=0;i<len2;i++)
        printf("%d ",mi[i]);
    printf("\n");
    for(int i=0;i<len1;i++)
        printf("%d ",mx[i]);
}