UESTC Training for Data Structures——J

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

/*维护一个不降的双向队列求最大值
  维护一个不增的双向队列求最小值*/

#include
#include
#include
#include
#define N 1000005
using namespace std;
int a[N],zhan[N];
int n,k;
void find_min()  //求最大值
{
    int head=0,tail=1;
    zhan[1]=1;
    for(int i=2;i<=k-1;i++)
    {
        if(a[i]<=a[zhan[head+1]]) { head=tail; zhan[++tail]=i; }
        else
        {
            while(tail!=head && a[zhan[tail]]>=a[i]) tail--;
            zhan[++tail]=i;
        }
    }
    for(int i=k;i<=n;i++)
    {
        if(tail-head==k) head++;
        if(a[i]<=a[zhan[head+1]]) { head=tail; zhan[++tail]=i; }
        else
        {
            while(tail!=head && a[zhan[tail]]>=a[i]) tail--;
            zhan[++tail]=i;
        }
        while(zhan[head+1]<=i-k) head++;
        printf("%d",a[zhan[head+1]]);
        if(i!=n) printf(" ");
    }
    printf("\n");
}
void find_max()  //求最小值
{
    int head=0,tail=1;
    zhan[1]=1;
    for(int i=2;i<=k-1;i++)
    {
        if(a[i]>=a[zhan[head+1]]) { head=tail; zhan[++tail]=i; }
        else
        {
            while(tail!=head && a[zhan[tail]]<=a[i]) tail--;
            zhan[++tail]=i;
        }
    }
    for(int i=k;i<=n;i++)
    {
        if(tail-head==k) head++;
        if(a[i]>=a[zhan[head+1]]) { head=tail; zhan[++tail]=i; }
        else
        {
            while(tail!=head && a[zhan[tail]]<=a[i]) tail--;
            zhan[++tail]=i;
        }
        while(zhan[head+1]<=i-k) head++;
        printf("%d",a[zhan[head+1]]);
        if(i!=n) printf(" ");
    }
    printf("\n");
}
int main()
{
    while(scanf("%d%d",&n,&k)!=EOF)
    {
        if(k>n) k=n;
        memset(a,0,sizeof(a));
        for(int i=1;i<=n;i++) scanf("%d",&a[i]);
        find_min();
        find_max();
    }
    return 0;
}