poj 2823 Sliding Window

最新推荐文章于 2021-07-29 16:00:21 发布

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#单调队列

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。通过双端队列维护窗口内的元素，确保快速更新最值。适用于数据流处理等场景。

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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 integersn 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

#include <iostream>
#include <cstdio>
#include <stack>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>
#include <set>
#include <deque>

using namespace std;

typedef long long ll;


int a[1000005];

struct node{
    int x;
    int v;
};
int n,m;

void a2(){
    deque<node> q;
    node no;
    no.v=a[1];
    no.x=1;
    q.push_back(no);

    for(int i=1;i<=n;i++){
        while(!q.empty()&&q.back().v>=a[i]){
            q.pop_back();
        }

        no.v=a[i];
        no.x=i;
        q.push_back(no);

        if(i>=m){
            if(i-q.front().x>=m){
                q.pop_front();
            }
            cout<<q.front().v<<" ";
        }
    }
    cout<<endl;
}

void a1(){
    deque<node> q;
    node no;
    no.v=a[1];
    no.x=1;
    q.push_back(no);

    for(int i=2;i<=n;i++){
        while(!q.empty()&&q.back().v<=a[i]){
            q.pop_back();
        }

        no.v=a[i];
        no.x=i;
        q.push_back(no);

        if(i>=m){
            if(i-q.front().x>=m){
                q.pop_front();
            }
            cout<<q.front().v<<" ";
        }
    }
    cout<<endl;
}

void solve(){

    scanf("%d %d",&n,&m);
    if(m==1){
        for(int i=1;i<=n;i++){
            scanf("%d",&a[i]);
        }
        for(int i=1;i<=n;i++){
            cout<<a[i]<<" ";
        }
        cout<<endl;
        for(int i=1;i<=n;i++){
            cout<<a[i]<<" ";
        }
        return;
    }
    for(int i=1;i<=n;i++){
        scanf("%d",&a[i]);
    }

    a2();
    a1();






}

int main(){
    solve();

    return 0;
}