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

    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:

    query: rec[1] rec[2] ... rec[K]
    

    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.

    Sample Input:

    12 3
    3 5 7 5 5 3 2 1 8 3 8 12
    

    Sample Output:

    5: 3
    7: 3 5
    5: 3 5 7
    5: 5 3 7
    3: 5 3 7
    2: 5 3 7
    1: 5 3 2
    8: 5 3 1
    3: 5 3 1
    8: 3 5 1
    12: 3 5 8

     1 #include<iostream>
     2 #include<vector>
     3 #include<set>
     4 using namespace std;
     5 struct Node{
     6   int qry, cnt;
     7   Node(){cnt=0;}
     8   bool operator<(const Node& node) const{
     9     if(this->cnt!=node.cnt) return this->cnt>node.cnt;
    10     return this->qry<node.qry;
    11   }
    12 };
    13 
    14 int main(){
    15   int n, k, i;
    16   scanf("%d%d", &n, &k);
    17   vector<Node> v(n+1);
    18   for(i=1; i<=n; i++) v[i].qry=i;
    19   int qry;
    20   scanf("%d", &qry);
    21   set<Node> s;
    22   v[qry].cnt=1;
    23   s.insert(v[qry]);
    24   for(i=1; i<n; i++){
    25     scanf("%d", &qry);
    26     set<Node>::iterator it = s.begin(), temp;
    27     printf("%d:", qry);
    28     bool flag = false;
    29     for(; it!=s.end(); it++){
    30         printf(" %d", (*it).qry);
    31         if((*it).cnt==v[qry].cnt && (*it).qry==qry){
    32             flag = true;
    33             temp = it;
    34         }
    35     }
    36     v[qry].cnt++;
    37     cout<<endl;
    38     if(flag) s.erase(temp);
    39     s.insert(v[qry]);
    40     if(s.size()>k) s.erase(--s.end());
    41   }
    42   return 0;
    43 }
  • 相关阅读:
    浅谈.NET下的多线程和并行计算系列文章索引
    浅谈.NET下的多线程和并行计算(六)线程池基础下
    浅谈.NET下的多线程和并行计算(八)Winform中多线程编程基础上
    项目优化经验——垃圾回收导致的性能问题
    浅谈.NET下的多线程和并行计算(五)线程池基础上
    站点静态资源优化合并解决方案
    浅谈.NET下的多线程和并行计算(二)线程基本知识
    浅谈.NET下的多线程和并行计算(十二)CLR via C#第三版阅读笔记(1)
    原谅我的说谎
    索爱手机GPRS的OTA设置[转]
  • 原文地址:https://www.cnblogs.com/mr-stn/p/9581623.html
Copyright © 2011-2022 走看看