1129 Recommendation System (25 分)

本文介绍了一个基于用户访问频率的简单推荐系统实现方案。系统通过跟踪用户对物品的访问次数,为用户推荐最常访问的物品。文章详细描述了输入输出规格,并提供了一个C++代码示例,展示了如何处理查询并生成推荐列表。

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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:

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

此题卡的一手好时,有点不太明白为何第3个测试点才160ms,不太明白为何当我vector开n大的时候就会 超时,而改成k大的话,就通过了,难道第三个测试点的测试数据n会超级大,以至于开着内存就超时了.........

#include<iostream>
#include<cstdio>
#include<map>
#include<algorithm>
#include<vector>
using namespace std;
int main(){
	int n,k,a;
	scanf("%d %d",&n,&k);
	vector<int> v(k,0);
	map<int,int> m;
	scanf("%d",&a);
	v[0]=a;
	m[a]++;
	for(int i = 1;i < n;i++){
		scanf("%d",&a);
		printf("%d:",a);
		for(int j = 0;j < k;j++){
			if(v[j])
			printf(" %d",v[j]);
		}
		printf("\n");
		m[a]++;
		int flag = 0;
		for(int j = 0;j < k;j++){
			if(v[j]==a){
				flag = 1;
				while(j!=0&&(m[a]>m[v[j-1]]||(m[a]==m[v[j-1]]&&a<v[j-1]))){
					swap(v[j],v[j-1]);
					j--;
				}
				break;
			}
		}
		if(flag == 0){
			if((m[a]>m[v.back()])||(m[a]==m[v.back()]&&a<v.back())){
				for(int j=0;j<k;j++){
                    if(m[a]>m[v[j]]||(m[a]==m[v[j]]&&a<v[j])){
                        v.insert(v.begin()+j,a);
                        v.pop_back();
                        break;
                    }
                }
			}
		}
	}
	return 0;
}

 

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