POJ 2823 Sliding Window

Posted 2020-08-06 勿忘初心0924

Sliding Window

Time Limit: 12000MS		Memory Limit: 65536K
Total Submissions: 54824		Accepted: 15777
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

Source

POJ Monthly--2006.04.28, Ikki

/*
第一道单调序列题，纪念

[题目大意]
给定一个长度为n的数列，求长度为k的定长连续子区间｛a1,a2,a3,a4,…,ak-1,ak｝{a2,a3,…,ak,ak+1}……中每个区间的最大值和最小值。
*/
#include<cstdio>
#include<iostream>
#include<cstring>
#define N 80050
#define LL long long
using namespace std;
LL q[N];//记录的是单调队列中元素在a数组中的下标
int n,k,a[N],I[N];
void getMax(){
    int head=1,tail=0;
    for(int i=1;i<k;i++){
        while(head<=tail&&a[q[tail]]<=a[i]) tail--;
        
        q[++tail]=i;
    }//现在1到k-1的优先队列先计算出来
    for(int i=k;i<=n;i++){
        while(head<=tail&&a[q[tail]]<=a[i])tail--;
        q[++tail]=i; 
        
        while(q[head]<=i-k)head++; //如果这个数不在当前区间内那么先不取这个数，再向当前区间选
        printf("%d ",a[q[head]]); 
    } 
} 
void getMin(){ 
    int head=1,tail=0;
    for(int i=1;i<k;i++){
        while(head<=tail&&a[q[tail]]>=a[i]) tail--;
        
        q[++tail]=i;
    }//现在1到k-1的优先队列先计算出来
    for(int i=k;i<=n;i++){
        while(head<=tail&&a[q[tail]]>=a[i])tail--;
        q[++tail]=i; 
        
        while(q[head]<=i-k)head++; //如果这个数不在当前区间内那么先不取这个数，再向当前区间选
        printf("%d ",a[q[head]]); 
    } 
} 

int main()
{
    //freopen("in.txt","r",stdin);
    while(scanf("%d%d",&n,&k)!=EOF&&n)
    {
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);
        getMax();
        printf("\n");
        getMin();
        printf("\n");
    }
    return 0;
}
/*
样例：
6 2
1 3 2 1 5 6

3 3 2 5 6
1 2 1 1 5

*/