Sliding Window

Time Limit : 24000/12000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 152   Accepted Submission(s) : 19

Problem 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 knumbers 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 -13 1 [3  -1  -3] 5  3  6  7 -33 1  3 [-1  -3  5] 3  6  7 -35 1  3  -1 [-3  5  3] 6  7 -35 1  3  -1  -3 [5  3  6] 7 36 1  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 integers <I>n</I> and <I>k</I> which are the lengths of the array and the sliding window. There are <I>n</I> integers in the second line. <br>

 

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. <br>

 

Sample Input

8 31 3 -1 -3 5 3 6 7

 

Sample Output

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

这题有毒啊，错的莫名奇妙

代码：

#include<iostream>#include<string>#include<cstdio>#include<algorithm>#include<cmath>#include<iomanip>#include<queue>#include<cstring>#include<map>using namespace std;int N,M,a[1000005],b1[1000005],v[1000005],S,mid,MM[1000005];void maxx(){    int i,j,k,l;    l=0;k=-1;j=0;    for(i=1;i<=M;i++)    {        while(k>=0&&a[i]>=b1[k]) k--;       b1[++k]=a[i];        v[k]=i;    }    MM[l++]=b1[0];    for(i=M+1;i<=N;i++)    {if(v[j]<i-M+1) j++;        while(k>=j&&a[i]>=b1[k]) k--;        b1[++k]=a[i];        v[k]=i;        MM[l++]=b1[j];    }    for(i=0;i<l-1;i++)        cout<<MM[i]<<" ";    cout<<MM[l-1]<<endl;}void minn(){    memset(v,0,sizeof(v));    memset(b1,0,sizeof(b1));    memset(MM,0,sizeof(MM));    int i,j,k,l;    l=0;k=-1;j=0;    for(i=1;i<=M;i++)    {        while(k>=0&&a[i]<=b1[k]) k--;       b1[++k]=a[i];        v[k]=i;    }    MM[l++]=b1[0];    for(i=M+1;i<=N;i++)    {        if(v[j]<i-M+1) j++;        while(k>=j&&a[i]<=b1[k]) k--;        b1[++k]=a[i];        v[k]=i;        MM[l++]=b1[j];    }    for(i=0;i<l-1;i++)        cout<<MM[i]<<" ";    cout<<MM[l-1]<<endl;}int main(){   int i;   while(cin>>N>>M)   {       for(i=1;i<=N;i++)        scanf("%d",&a[i]);       minn();       maxx();   }   return 0;}

我的代码：

#include<iostream>#include<string>#include<cstdio>#include<algorithm>#include<cmath>#include<iomanip>#include<queue>#include<cstring>#include<map>using namespace std;#define M 1000005int n,k,a[M],q[M],id[M],ans[M];void get_min(){    int i,head=1,tail=0;    for(i=1;i<k;i++)    {        while(head<=tail&&a[i]<=q[tail])tail--;        q[++tail]=a[i];        id[tail]=i;    }    for(i=k;i<=n;i++)    {        if(id[head]<=i-k) head++;        while(head<=tail&&a[i]<=q[tail])tail--;        q[++tail]=a[i];        id[tail]=i;        ans[i]=q[head];    }    for(i=k;i<n;i++)    printf("%d ",ans[i]);    printf("%d\n",ans[n]);}void get_max(){    int i,head=1,tail=0;    for(i=1;i<k;i++)    {        while(head<=tail&&a[i]>=q[tail])tail--;        q[++tail]=a[i];        id[tail]=i;    }    for(i=k;i<=n;i++)    {        if(id[head]<=i-k) head++;        while(head<=tail&&a[i]>=q[tail])tail--;        q[++tail]=a[i];        id[tail]=i;        ans[i]=q[head];    }    for(i=k;i<n;i++)    printf("%d ",ans[i]);    printf("%d\n",ans[n]);}int main(){    while(scanf("%d%d",&n,&k)!=EOF) {        for(int i=1;i<=n;i++)            scanf("%d",&a[i]);        get_min();        get_max();    }    return 0;}