zoukankan      html  css  js  c++  java
  • POJ2823 Sliding Window(单调队列)

    单调队列,我用deque维护。这道题不难写,我第二次写单调队列,1次AC。

    -----------------------------------------------------------------------------------

    #include<cstdio>
    #include<cstring>
    #include<deque>
    #define rep(i,r) for(int i=0;i<r;i++)
    #define clr(x,c) memset(x,c,sizeof(x))
    #define Rep(i,l,r) for(int i=l;i<r;i++)
    using namespace std;
    const int maxn=1000000+5;
    int n,k;
    int Min[maxn],Max[maxn];
    deque<int> minq,maxq;
    deque<int> minnum,maxnum;
    int main()
    {
    // freopen("test.in","r",stdin);
    // freopen("test.out","w",stdout);
    while(scanf("%d%d",&n,&k)==2) {
    int t;
    rep(i,n) {
    scanf("%d",&t);
    if(minq.empty()) {
    minq.push_back(t); minnum.push_back(i);
    } else {
    while(!minq.empty() && minnum.front()<=i-k) {
       minq.pop_front();
       minnum.pop_front();
       }
       while(!minq.empty() && minq.back()>=t) {
           minq.pop_back();
           minnum.pop_back();
       }
       minq.push_back(t); minnum.push_back(i);
    }
    if(maxq.empty()) {
    maxq.push_back(t); maxnum.push_back(i);
    } else {
    while(!maxq.empty() && maxnum.front()<=i-k) {
       maxq.pop_front();
       maxnum.pop_front();
       }
       while(!maxq.empty() && maxq.back()<=t) {
           maxq.pop_back();
           maxnum.pop_back();
       }
       maxq.push_back(t); maxnum.push_back(i);
    }
    Min[i]=minq.front(); Max[i]=maxq.front();    
       }
       Rep(i,k-1,n) printf("%d ",Min[i]);
    printf(" ");
    Rep(i,k-1,n) printf("%d ",Max[i]);
    printf(" ");
    }
    return 0;
    }

    ----------------------------------------------------------------------------------- 

    Sliding Window
    Time Limit: 12000MSMemory Limit: 65536K
    Total Submissions: 41760Accepted: 12360
    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 -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 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

  • 相关阅读:
    可编辑的div -> 编辑框
    图片的缩放处理
    枚举转换成字典集合的通用方法
    SQLPrompt_9.5.5.9830破解版含注册机_永久激活不掉线
    配置阿里云docker镜像地址
    Virtualbox设置固定IP
    史上最全的 vi 命令大全,建议手册
    ASP.NET Core WebApi使用Swagger生成api说明文档看这篇就够了
    vue Slot理解
    windows上pip安装及使用详解
  • 原文地址:https://www.cnblogs.com/JSZX11556/p/4354720.html
Copyright © 2011-2022 走看看