1129 Recommendation System (25 分)(set排序的运用)

题目描述:

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
思路:题目的意思是让我们从左到右扫描过去,找出出现次序前k大的数,我们可以运用set容器,修改排序方式,维护set中的数,及时删除这个数的上个版本,并加入当前版本,即num++后
AC代码:

#include<iostream>
#include<string>
#include<map>
#include<set>
#include<algorithm>
#include<queue>
#include<cmath>
#include<string.h>
#include<unordered_map>
#include<unordered_set>
using namespace std;
#define INT_MIN -2147483648
#define INT_MAX 2147483647
const int maxn = 5e4 + 1;
int read() {
    char p;
    while ((p = getchar()) < '0' || p > '9');
    int sum = p - 48;
    while ((p = getchar()) >= '0' && p <= '9')sum = sum * 10 + p - '0';
    return sum;
}
int n, k;
struct node {
    int cnt, id;
    bool operator<(const node &b)const {
        return cnt != b.cnt ? cnt > b.cnt : id < b.id;
    }
};
int num[maxn];
int main() {
    //freopen("test.txt", "r", stdin);
    n = read(), k = read();
    set<node>S;
    for (int i = 0; i < n; i++) {
        int tmp = read();
        if (i != 0) {
            printf("%d:", tmp);
            int m = 0;
            for (auto it = S.begin(); it != S.end() && m < k;m++,it++) {
                printf(" %d", it->id);
            }
            printf("\n");
        }
        auto it = S.find({ num[tmp],tmp });//删除tmp在set中的前一个版本
        if (it != S.end())S.erase(it);//如果没找到说明是第一次插入
        num[tmp]++;
        S.insert({ num[tmp],tmp });//插入tmp的新版本
    }
    return 0;
}