Sliding Window

Sliding Window

Sliding Window

Time Limit: 12000MS		Memory Limit: 65536K
Total Submissions: 7213		Accepted: 1859
Case Time Limit: 5000MS

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

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
这题让我想到了 几个月前的浙大月赛题的区间最大最小值，用队列维护的方法还是不会
用线段树AC，有点郁闷，我的半吊子线段树啊   居然10s才ac，不过幸好服务器没挂。。。。。。
代码如下
 

#include < stdio.h >
#include < string .h >
#define  MAX 1001010
struct  node {
    int l,r;
    int m;
} ;
node Max_Stree[ 2 * MAX];
node Min_Stree[ 2 * MAX];
int  w[MAX];
int  getmax( int  a, int  b)
{
    return a>b?a:b;    
}
int  getmin( int  a, int  b)
{
    return a>b?b:a;    
}
int  Build_Max( int  now, int  l, int  r) {
    Max_Stree[now].l=l;
    Max_Stree[now].r=r;
    if(l==r)Max_Stree[now].m=w[l];
    else {
        int mid=(l+r)>>1;
        int max1=Build_Max(2*now,l,mid);
        int max2=Build_Max(2*now+1,mid+1,r);
        Max_Stree[now].m=getmax(max1,max2);    
    }
    return Max_Stree[now].m;    
}
int  Build_Min( int  now, int  l, int  r) {
    Min_Stree[now].l=l;
    Min_Stree[now].r=r;
    if(l==r)Min_Stree[now].m=w[l];
    else {
        int mid=(l+r)>>1;
        int min1=Build_Min(2*now,l,mid);
        int min2=Build_Min(2*now+1,mid+1,r);
        Min_Stree[now].m=getmin(min1,min2);    
    }
    return Min_Stree[now].m;    
}
int  Find_Max( int  now, int  l, int  r) {
    int left=Max_Stree[now].l;
    int right=Max_Stree[now].r;
    if(left==l&&right==r)
        return Max_Stree[now].m;
    int mid=(left+right)>>1;
    if(mid+1>r)return Find_Max(2*now,l,r);
    if(mid<l)return Find_Max(2*now+1,l,r);    
    else return getmax(Find_Max(2*now,l,mid),Find_Max(2*now+1,mid+1,r));
}
int  Find_Min( int  now, int  l, int  r) {
    int left=Min_Stree[now].l;
    int right=Min_Stree[now].r;
    if(left==l&&right==r)
        return Min_Stree[now].m;
    int mid=(left+right)>>1;
    if(mid+1>r)return Find_Min(2*now,l,r);
    if(mid<l)return Find_Min(2*now+1,l,r);    
    else return getmin(Find_Min(2*now,l,mid),Find_Min(2*now+1,mid+1,r));
}