poj_2823(单调队列)

最新推荐文章于 2021-01-25 20:02:52 发布

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最新推荐文章于 2021-01-25 20:02:52 发布

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分类专栏： POJ 数据结构∨ 文章标签： C++ 基础算法单调队列

本文链接：https://blog.youkuaiyun.com/u012198382/article/details/38377903

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本文介绍如何使用单调队列解决区间最值问题，包括输入格式、输出要求及代码实现，通过实例演示算法应用。

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Sliding Window

Time Limit: 12000MS		Memory Limit: 65536K
Total Submissions: 38520		Accepted: 11415
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

一道裸的单调队列，求区间最值问题。线段树8秒多过，单调队列4秒多过。可作为单调队列的学习题目。

先说一下单调队列是一种什么样的队列。单调队列分为递增和递减队列，下面以递增队列为例：

1.该队列时刻保持队头元素最小且递增。元素从队尾插入，如果小于队尾元素，那么就删去队尾元素，再插入。

2.同时需要一个数组保存元素下标，以本题为例，区间长度为k,队头元素的下标小于当前元素下标i-k+1时，也就是此时队头元素已经不在k区间内了，就把队头元素删去。

#include<iostream>
#include<cstdio>
#define M 1000001
using namespace std;

int n,k;
int a[M];
int q[M];
int p[M];

void get_min()
{
    int head=1;
    int tail=0;
    for(int i=0;i<k-1;i++)
    {
        while(head<=tail&&a[i]<=q[tail])
            --tail;
        q[++tail]=a[i];
        p[tail]=i;
    }
    for(int i=k-1;i<n;i++)
    {
        while(head<=tail&&a[i]<=q[tail])
            --tail;
        q[++tail]=a[i];
        p[tail]=i;
        while(p[head]<i-k+1)
        {
            ++head;
        }
        cout<<q[head]<<" ";
    }
    cout<<endl;
}

void get_max()
{
    int head=1;
    int tail=0;
    for(int i=0;i<k-1;i++)
    {
        while(head<=tail&&a[i]>=q[tail])
            --tail;
        q[++tail]=a[i];
        p[tail]=i;
    }
    for(int i=k-1;i<n;i++)
    {
        while(head<=tail&&a[i]>=q[tail])
            --tail;
        q[++tail]=a[i];
        p[tail]=i;
        while(p[head]<i-k+1)
        {
            ++head;
        }
        cout<<q[head]<<" ";
    }
    cout<<endl;
}

int main()
{
    //freopen("d:\\test.txt","r",stdin);
    scanf("%d%d",&n,&k);
    for(int i=0;i<n;i++)
    {
        scanf("%d",&a[i]);
    }
    get_min();
    get_max();
    return 0;
}