POJ

滑动窗口的最值问题

http://poj.org/problem?id=2823

链接：https://ac.nowcoder.com/acm/problem/51001
来源：牛客网

时间限制：C/C++ 1秒，其他语言2秒
空间限制：C/C++ 32768K，其他语言65536K
64bit IO Format: %lld

题目描述

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. 

输入描述:

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.

输出描述:

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.

示例1

输入

8 3
1 3 -1 -3 5 3 6 7

输出

-1 -3 -3 -3 3 3
3 3 5 5 6 7

#include <bits/stdc++.h>
typedef long long LL;
#define pb push_back
#define mst(a) memset(a,0,sizeof(a))
const int INF = 0x3f3f3f3f;
const double eps = 1e-8;
const int mod = 1e9+7;
const int maxn = 1e6+10;
using namespace std;

int a[maxn];
vector<int> ans1,ans2;
deque<int> qe1,qe2;

int main()
{
    #ifdef DEBUG
    freopen("sample.txt","r",stdin); //freopen("data.out", "w", stdout);
    #endif
    
    int n,m;
    scanf("%d %d",&n,&m);
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
    for(int i=1;i<=n;i++)
    {
        //维护最小值队列qe1
        if(!qe1.empty()&&qe1.front()<=i-m) qe1.pop_front();
        if(qe1.empty()) qe1.push_back(i);
        else
        {
            if(a[i]<=a[qe1.front()])
            {
                qe1.clear();
                qe1.push_back(i);
            }
            else
            {
                if(qe1.size()==1) qe1.push_back(i);
                else
                {
                    if(a[i]>=a[qe1.back()]) qe1.push_back(i);
                    else
                    {
                        int t = qe1.back();
                        while(a[i]<a[t])
                        {
                            qe1.pop_back();
                            t = qe1.back();
                        }
                        qe1.push_back(i);
                    }
                }
            }
        }
        //维护最大值队列qe2
        if(!qe2.empty()&&qe2.front()<=i-m) qe2.pop_front();
        if(qe2.empty()) qe2.push_back(i);
        else
        {
            if(a[i]>=a[qe2.front()])
            {
                qe2.clear();
                qe2.push_back(i);
            }
            else
            {
                if(qe2.size()==1) qe2.push_back(i);
                else
                {
                    if(a[i]<=a[qe2.back()]) qe2.push_back(i);
                    else
                    {
                        int t = qe2.back();
                        while(a[i]>a[t])
                        {
                            qe2.pop_back();
                            t = qe2.back();
                        }
                        qe2.push_back(i);
                    }
                }
            }
        }
        
        if(i>=m)
        {
            ans1.push_back(a[qe1.front()]);
            ans2.push_back(a[qe2.front()]);
        }
    }

    for(int i=0;i<=n-m;i++)
        printf(i==n-m?"%d
":"%d ",ans1[i]);
    for(int i=0;i<=n-m;i++)
        printf(i==n-m?"%d
":"%d ",ans2[i]);
    
    return 0;
}

-