zoukankan      html  css  js  c++  java
  • 1129 Recommendation System

    1129 Recommendation System (25 分)

    Recommendation system predicts the preference that a user would give to an item. Now you are asked to program a very simple recommendation system that rates the user's preference by the number of times that an item has been accessed by this user.

    Input Specification:

    Each input file contains one test case. For each test case, the first line contains two positive integers: N (≤ 50,000), the total number of queries, and K (≤ 10), the maximum number of recommendations the system must show to the user. Then given in the second line are the indices of items that the user is accessing -- for the sake of simplicity, all the items are indexed from 1 to N. All the numbers in a line are separated by a space.

    Output Specification:

    For each case, process the queries one by one. Output the recommendations for each query in a line in the format:

    where query is the item that the user is accessing, and rec[i] (i=1, ... K) is the i-th item that the system recommends to the user. The first K items that have been accessed most frequently are supposed to be recommended in non-increasing order of their frequencies. If there is a tie, the items will be ordered by their indices in increasing order.

    Note: there is no output for the first item since it is impossible to give any recommendation at the time. It is guaranteed to have the output for at least one query.

    知识点:STL的使用,排序

    思路:每一次输入后,都要让所有元素有序。如果利用线性结构,那么他的时间复杂度是O(n),在数据量大的时候也是不行的。利用STL的set,插入的时间复杂度降为O(logN)

     1 #include <cstdio>
     2 #include <set>
     3 using namespace std;
     4 const int maxn = 500050;
     5 
     6 int clickTime[maxn];
     7 struct nodetype{
     8     int v,cnt;
     9     nodetype(int a,int b) : v(a), cnt(b) {}
    10     bool operator < (const nodetype &a) const    {
    11         return (cnt!=a.cnt)?cnt>a.cnt:v<a.v;
    12     }    
    13 };
    14 
    15 int main(int argc, char *argv[]) {
    16     fill(clickTime,clickTime+maxn,0);
    17     
    18     int n,k,tmp;
    19     scanf("%d %d",&n,&k);
    20     set<nodetype> s;
    21     for(int i=1;i<=n;i++){
    22         scanf("%d",&tmp);
    23         if(i!=1){
    24             printf("%d:",tmp);
    25             int cnt=0;
    26             for(auto it=s.begin();cnt<k&&it!=s.end();it++){
    27                 printf(" %d",it->v);
    28                 cnt++;
    29             }
    30             printf("
    ");
    31         }
    32         
    33         auto it = s.find(nodetype(tmp,clickTime[tmp]));
    34         if(it!=s.end()){
    35             s.erase(nodetype(tmp,clickTime[tmp]));
    36         }
    37         s.insert(nodetype(tmp,clickTime[tmp]+1));
    38         clickTime[tmp]++;
    39     }
    40 }
  • 相关阅读:
    Linux0.11内核--fork进程分析
    Linux0.11内核--内存管理之1.初始化
    Linux0.11内核--进程调度分析之2.调度
    Linux0.11内核--进程调度分析之1.初始化
    github
    推荐大家一个靠谱的论文检测平台。重复的部分有详细出处以及具体修改意见,能直接在文章上做修改,全部改完一键下载就搞定了。他们现在正在做毕业季活动, 赠送很多免费字数,可以说是十分划算了!地址是:https://www.paperpass.com/
    妈妈再也不用担心我找idea激活码了
    eclipse集成tomcat
    DNS--localhost
    RFC 2819)第5节"Definitions"除外的全部内容
  • 原文地址:https://www.cnblogs.com/lokwongho/p/9908526.html
Copyright © 2011-2022 走看看