zoukankan      html  css  js  c++  java
  • POJ2823 ·Sliding Window

    传送门

    Sliding Window
    Time Limit: 12000MS   Memory Limit: 65536K
         
    Case Time Limit: 5000MS

    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 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 positionMinimum valueMaximum 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

     
    又刷水题了T^T..果果的单调队列。。或者别的数据结构乱搞。。不过线段树什么的能不能过我就不晓得了。。
     1 #include<set>
     2 #include<queue>
     3 #include<cstdio>
     4 #include<cstdlib>
     5 #include<cstring>
     6 #include<iostream>
     7 #include<algorithm>
     8 using namespace std;
     9 const int N = 1000007;
    10 #define For(i,n) for(int i=1;i<=n;i++)
    11 #define Rep(i,l,r) for(int i=l;i<=r;i++)
    12 
    13 int n,Loc[N],k,Q[N],A[N]; 
    14 
    15 void Max(){    
    16     int l = 1 ,r = 0;
    17     For(i,k-1){
    18         while(l<=r&&Q[r]<=A[i]) r--;
    19         Q[++r]=A[i];Loc[r]=i;
    20     }
    21     Rep(i,k,n){
    22         while(l<=r&&Q[r]<=A[i]) r--;
    23         Q[++r]=A[i];Loc[r]=i;
    24         while(Loc[l]<=i-k) l++;                    
    25         if(k==n)  printf("%d
    ",Q[l]);
    26         else      printf("%d ",Q[l]);
    27     } 
    28 }
    29 
    30 void Min(){
    31     int l = 1 ,r = 0;
    32     For(i,k-1){
    33         while(l<=r&&Q[r]>=A[i]) r--;
    34         Q[++r]=A[i];Loc[r]=i;
    35     }
    36     Rep(i,k,n){
    37         while(l<=r&&Q[r]>=A[i]) r--;
    38         Q[++r]=A[i];Loc[r]=i;
    39         while(Loc[l]<=i-k) l++;                   
    40         if(k==n)  printf("%d
    ",Q[l]);
    41         else      printf("%d ",Q[l]);
    42     } 
    43     puts("");
    44 }
    45 
    46 int main(){
    47     scanf("%d%d",&n,&k);
    48     For(i,n) scanf("%d",&A[i]);
    49     Min();Max();
    50     return 0;
    51 }
  • 相关阅读:
    Linq to Sql 总生成 where ID is null 的解决办法
    jexus asp.net Linux Web Server
    怎么修改盘符
    TCP、UDP、IP 协议分析
    和浙大妹子聊准备笔面
    2015腾讯笔试大题
    导出/打印项目数据报表需要设置IE浏览器
    Eclipse快捷键
    Eclipse如何解决启动慢?
    eclipse mars4.5安装hibernate开发环境
  • 原文地址:https://www.cnblogs.com/zjdx1998/p/3948054.html
Copyright © 2011-2022 走看看