PAT 1129 Recommendation System[比较]

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

 题目大意：推荐系统，有n个查询，从第二个查询开始，给出每个可能的结果，结果根据前面查询次数最多的输出，如果查询次数相同，那么根据序号升序输出。

 //我能想到的就是，对一个结构体数组，每次遇到就+1，并且排序sort按次数和序号，对每一个预测时都输出，这明显是会超时的。

代码来自： https://www.liuchuo.net/archives/3848

#include <iostream>
#include <set>
using namespace std;
int book[50001];
struct node {
    int value, cnt;
    bool operator < (const node &a) const {//重载运算符，学习了，参数传进来的是
        //另外一个和当前的比较。
        return (cnt != a.cnt) ? cnt > a.cnt : value < a.value;
    }
};
int main() {
    int n, k, num;
    scanf("%d%d", &n, &k);
    set<node> s;
    for (int i = 0; i < n; i++) {
        scanf("%d", &num);
        if (i != 0) {
            printf("%d:", num);
            int tempCnt = 0;
            for(auto it = s.begin(); tempCnt < k && it != s.end(); it++) {
                printf(" %d", it->value);//使用迭代器来遍历set,并使用it->value进行输出结构体属性值，学习了。
                tempCnt++;
            }
            printf("
");
        }
        auto it = s.find(node{num, book[num]});//原来在set里查找结构体也可以这样。
        //如果原来出现了，将原来的删去，
        if (it != s.end()) s.erase(it);
        book[num]++;//并且插入进更新后的点。
        s.insert(node{num, book[num]});
    }
    return 0;
}

1.利用了set的自排序特性，set中存储了结构体，给结构体重载了运算符函数，首先按出现次数排序，再按号码排序，学习了。

2.只进行一次遍历，i!=0时就开始输出，根据当前set的大小和k进行输出，学习了！

//真厉害，好好再想一下这道题。