Problem J

Problem Description

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

There are several cases in the input file, each case 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 for each input. 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<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>

using namespace std;
#define N 1000005
int a[N];
int j;
int M[N];
int Ma;
int MI[N];
int mina;

int min(int x,int y)
{
    if(x<y)return x ;
    else return y;
}

int max(int x,int y)
{
    if(x>y)return x;
    else return y ;
}

struct node 
{
    int l;
    int r;
    int ma;
    int mi;
}t[N*4];

void make (int sl,int sr,int i)
{
    t[i].l=sl;
    t[i].r=sr;
    if(sl==sr)
    {
        t[i].ma=a[sr];
        t[i].mi=a[sr];
    }
    else 
    {
        int mid=(t[i].l+t[i].r)/2;
        make (sl,mid,i*2);
        make (mid+1,sr,i*2+1);
        t[i].ma=max(t[i*2].ma,t[i*2+1].ma);        
        t[i].mi=min(t[i*2].mi,t[i*2+1].mi);
    }
}
    
int  query1(int st,int ed,int i)
{
    if(st==t[i].l&&ed==t[i].r)
    //    M[j++]=t[i].ma;
        return t[i].ma;
    else if(t[i].l<t[i].r)
    {
        int mid=(t[i].l+t[i].r)/2;
        if(st>mid)
            return query1(st,ed,i*2+1);
        else if(ed<=mid)
            return query1(st,ed,i*2);
        else 
        {
            return max(query1(st,mid,i*2),query1(mid+1,ed,i*2+1));
        }
    }
}

int  query2(int st,int ed,int i)
{
    if(st==t[i].l&&ed==t[i].r)
        return t[i].mi;
    else if(t[i].l<t[i].r)
    {
        int mid=(t[i].l+t[i].r)/2;
        if(st>mid)
            return query2(st,ed,i*2+1);
        else if(ed<=mid)
            return query2(st,ed,i*2);
        else 
        {
            return min(query2(st,mid,i*2),query2(mid+1,ed,i*2+1));
        }
    }
}

int main()
{
    int n,m;
    int e;
    int i;
    while(scanf("%d %d",&n,&m)!=EOF)
    {
        memset(M,0,sizeof(M));
        memset(MI,0,sizeof(MI));
        for(i=1;i<=n;i++)
            scanf("%d",&a[i]);
        make(1,n,1);
        j=0;
        for(i=1;i<n&&(i+m-1)<=n;i++)
        {
            M[j++]=query1(i,i+m-1,1);
        }
        e=0;
        for(i=1;i<n&&(i+m-1)<=n;i++)
        {
            MI[e++]=query2(i,i+m-1,1);
        }
        for(i=0;i<e-1;i++)
            printf("%d ",MI[i]);
        printf("%d\n",MI[i]);
        for(i=0;i<e-1;i++)
            printf("%d ",M[i]);
        printf("%d\n",M[i]);
    }
    return 0;
}