[POJ2823]Sliding Window(单调队列)

[POJ2823]Sliding Window(单调队列)

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

题意简化

给定一个长度为n的数列，从左至右输出每个长度为m的数列段内的最小数和最大数。n,m<=10^6

这题是单调队列的模板题，至于单调队列我写了个博客大家可以看看：
http://blog.youkuaiyun.com/cabi_zgx/article/details/52701266

我还写了个ppt那是我在学校讲题用的，给个链接吧，不用钱大家放心下载：
http://download.youkuaiyun.com/detail/cabi_zgx/9650649

C++code:

#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
struct node
{
    int x,y;
}v[1110000];
int a[1110000],mn[1110000],mx[1110000];
int n,m;
void getmin()
{
    int i;
    int head=1,tail=0;
    for(i=1;i<m;i++)
    {
        while(head<=tail&&v[tail].x>=a[i])tail--;
        tail++;v[tail].x=a[i];v[tail].y=i;
    }
    for(;i<=n;i++)
    {
        while(head<=tail&&v[tail].x>=a[i])tail--;
        tail++;v[tail].x=a[i];v[tail].y=i;
        while(v[head].y<i-m+1)head++;
        mn[i-m+1]=v[head].x;
    }
}
void getmax()
{
    int i;
    int head=1,tail=0;
    for(i=1;i<m;i++)
    {
        while(head<=tail&&v[tail].x<=a[i])tail--;
        tail++;v[tail].x=a[i];v[tail].y=i;
    }
    for(;i<=n;i++)
    {
        while(head<=tail&&v[tail].x<=a[i])tail--;
        tail++;v[tail].x=a[i];v[tail].y=i;
        while(v[head].y<i-m+1)head++;
        mx[i-m+1]=v[head].x;
    }
}
int main()
{
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)scanf("%d",&a[i]);
    getmin();getmax();
    printf("%d",mn[1]);
    for(int i=2;i<=n-m+1;i++)printf(" %d",mn[i]);printf("\n");
    printf("%d",mx[1]);
    for(int i=2;i<=n-m+1;i++)printf(" %d",mx[i]);printf("\n");
    return 0;
}