POJ 2823 Sliding Window-白红宇

POJ 2823 Sliding Window

阅读量：5352 次

发布时间：2019-06-15

本文共 3813 字，大约阅读时间需要 12 分钟。

Sliding Window

Time Limit: 12000MS		Memory Limit: 65536K
Total Submissions: 26178		Accepted: 7738
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

kis 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 31 3 -1 -3 5 3 6 7

Sample Output

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

Source

, Ikki

//开始有线段树做 9188MS

//后面改用刘老师的基础数据结构里面介绍的用单调队列做 5313MS

//不过用线段树写的不止简洁一丁点、

//也有可能是偶代码技术还不好

#include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           #include 
           
            #define N 1000004#define lson l,m,k<<1#define rson m+1,r,k<<1|1using namespace std;struct node{ int min,max;};node st[N<<2];void build(int l,int r,int k){ if(l==r) { scanf("%d",&st[k].min); st[k].max=st[k].min; return ; } int m=(l+r)>>1; build(lson); build(rson); st[k].max=max(st[k<<1].max,st[k<<1|1].max); st[k].min=min(st[k<<1].min,st[k<<1|1].min);}int quMin(int &L,int &R,int l,int r,int k){ if(L<=l&&R>=r) { return st[k].min; } int m=(l+r)>>1; int t1=1000000000,t2=1000000000; if(L<=m) t1=quMin(L,R,lson); if(R>m) t2=quMin(L,R,rson); return t1
            
             =r) { return st[k].max; } int m=(l+r)>>1; int t1=-1000000000,t2=-1000000000; if(L<=m) t1=quMax(L,R,lson); if(R>m) t2=quMax(L,R,rson); return t1>t2?t1:t2;}int main(){ int i,k; int n,m,r; while(scanf("%d%d",&n,&k)!=EOF) { build(1,n,1); m=n-k; for(i=1;i<=m;i++) r=i+k-1,printf("%d ",quMin(i,r,1,n,1)); r=m+1; printf("%d\n",quMin(r,n,1,n,1)); for(i=1;i<=m;i++) r=i+k-1,printf("%d ",quMax(i,r,1,n,1)); r=m+1; printf("%d\n",quMax(r,n,1,n,1)); } return 0;}

#include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           #define N 1000004using namespace std;struct node{ int id,val;};node M[N];int front,back;int a[N];int main(){ int n,k,i; while(scanf("%d%d",&n,&k)!=EOF) { front=back=1; scanf("%d",&a[1]); M[front].val=a[1];M[front].id=1; for(i=2;i<=k;i++) { scanf("%d",&a[i]); if(a[i]<=M[back].val) { while(1) { M[back].val=a[i],M[back].id=i; if(back>front&&a[i]<=M[back-1].val) back--; else break; } } else { M[++back].val=a[i]; M[back].id=i; } } printf("%d",M[front].val); for(i=k+1;i<=n;i++) { scanf("%d",&a[i]); if(a[i]<=M[back].val) { while(1) { M[back].val=a[i],M[back].id=i; if(back>front&&a[i]<=M[back-1].val) back--; else break; } } else { M[++back].val=a[i]; M[back].id=i; } while(M[front].id
           
            =M[back].val) { while(1) { M[back].val=a[i],M[back].id=i; if(back>front&&a[i]>=M[back-1].val) back--; else break; } } else { M[++back].val=a[i]; M[back].id=i; } } printf("%d",M[front].val); for(i=k+1;i<=n;i++) { if(a[i]>=M[back].val) { while(1) { M[back].val=a[i],M[back].id=i; if(back>front&&a[i]>=M[back-1].val) back--; else break; } } else { M[++back].val=a[i]; M[back].id=i; } while(M[front].id