POJ 2823——Sliding Window

Sliding Window

Time Limit: 12000MS Memory Limit: 65536KTotal Submissions: 24525 Accepted: 7206Case Time Limit: 5000MS

Description

An array of size n ≤ 10⁶ is given to you. There is a sliding window of sizek which is moving from the very left of the array to the very right. You can only see thek 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 positionMinimum valueMaximum value[1  3  -1] -3  5  3  6  7-131 [3  -1  -3] 5  3  6  7-331  3 [-1  -3  5] 3  6  7-351  3  -1 [-3  5  3] 6  7-351  3  -1  -3 [5  3  6] 7361  3  -1  -3  5 [3  6  7]37

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 integersn and k which are the lengths of the array and the sliding window. There aren 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 31 3 -1 -3 5 3 6 7

Sample Output

-1 -3 -3 -3 3 33 3 5 5 6 7

Source

POJ Monthly--2006.04.28, Ikki

/*维护一个不增的双向队列求最大值  维护一个不降的双向队列求最小值*/#include<iostream>#include<cstring>#include<cmath>#include<cstdio>#define N 1000005using namespace std;int a[N],zhan[N];int n,k;void find_min()  //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");}void find_max()  //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");}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;}