PAT 1129. Recommendation System (25)

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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:

1
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

这个题目做起来还是有点艰辛的，想了好久。 一开始用的map，超时了。后来用set，插入的时候先删除，AC了。主要是用map不能直接按值排序，变相的方法是放在一个线性的容器中，用sort或者qsort去排序，依然超时。用set，插入的时候，先查找元素是否存在，存在则删除，插入一个新的值。这样构造出来的set是按元素出现的次数排好序的，可以直接输出set元素的前k(k > set.size)个，否则就输出set的全部元素。

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#include <cstdio>
#include <set>
#include <vector>

using namespace std;

struct node {
    int item, cnt;

    node(int a, int b) : item(a), cnt(b) {}

    bool operator<(const struct node a) const {
        if (cnt > a.cnt) return true;
        if (cnt < a.cnt) return false;
        if (item < a.item) return true;
        return false;
    }
};

set<node> query;
vector<int> times(50010);

void printQuery(int k) {
    int i = 0;
    for (auto it = query.begin(); i < k && it != query.end(); it++, i++) {
        printf(" %d", it->item);
    }
    printf("\n");
}

int main() {
    int n = 0, k = 0, pre = 0, item = 0;;
    scanf("%d %d %d", &n, &k, &pre);
    for (int i = 1; i < n; i++) {
        scanf("%d", &item);
        printf("%d:", item);
        auto f = query.find(node(pre, times[pre]));
        if (f != query.end()) {
            query.erase(f);
        }
        times[pre]++;
        query.insert(node(pre, times[pre]));
        printQuery(k);
        pre = item;
    }
    return 0;
}