PAT (Advanced Level) Practise 1129 Recommendation System (25)

1129. Recommendation System (25)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 (<= 50000), 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 33 5 7 5 5 3 2 1 8 3 8 12

Sample Output:

5: 37: 3 55: 3 5 75: 5 3 73: 5 3 72: 5 3 71: 5 3 28: 5 3 13: 5 3 18: 3 5 112: 3 5 8

题意：根据用户每次点击的东西的编号，输出k个他在点当前编号之前应该给这个用户推荐的商品的编号，也就是输出用户曾经点击过的商品编号的最多的前k个，如果恰好两个商品有相同的点击次数，就输出编号较小的那个

解题思路：用set来统计

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <cmath>#include <map>#include <set>#include <stack>#include <queue>#include <vector>#include <bitset>#include <functional>using namespace std;#define LL long longconst int INF = 0x3f3f3f3f;int n, k,a,visit[50005];struct node{int id;int sum;friend bool operator <(node a, node b){if (a.sum != b.sum) return a.sum > b.sum;else return a.id < b.id;}}pre;int main(){while (~scanf("%d%d%d", &n, &k, &a)){memset(visit, 0, sizeof visit);pre.id = a, pre.sum = 1;visit[a]++;set<node>s;s.insert(pre);for (int i = 1; i < n; i++){scanf("%d", &a);printf("%d:", a);set<node>::iterator it = s.begin();for(int j=1;j<=k;j++){if (it == s.end()) break;printf(" %d", it->id);it++;}printf("\n");pre.id = a;pre.sum = visit[a];if (visit[a]++) s.erase(pre);pre.id = a; pre.sum = visit[a];s.insert(pre);}}return 0;}