PAT甲级1129解题报告

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本文介绍了一个基于用户访问频率的简易推荐系统实现方案。系统通过跟踪用户对物品的访问次数，实时推荐用户可能感兴趣的前K个物品。使用了map和自定义结构体配合set的数据结构，实现了物品按访问频率及索引排序的高效推荐。

1129 Recommendation System （25 point(s)）

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:

题目大意：给一堆数，在每次给出一个数后查询前面的数出现次数最多的K个数，不满K个数就输出所有数，第一个数不查询，因为没有数。

解题思路：一开始用个map和vector排序，没过，后来用了个set，简单的自定义一个结构体，规定<方法，利用set自动排序的特性自动排序就过了。

#include<iostream>
#include<string.h>
#include<vector>
#include<algorithm>
#include<iomanip>
#include<time.h>
#include<math.h>
#include<set>
#include<list>
#include<climits>
#include<queue>
#include<cstring>
#include<map>
#include<stack>
#include<string>
using namespace std;
map<int, int> a;
struct node {
	int v;
	int c;
	node(int v, int c) {
		this->v = v;
		this->c = c;
	}
	bool operator<(const struct node & tmp)const {
		if (this->c > tmp.c) {
			return true;
		}
		else if (this->c == tmp.c) {
			return this->v < tmp.v;
		}
		else {
			return false;
		}
	}
};
set<node> cur;
int main()
{
	int N, K;
	scanf("%d %d", &N, &K);
	for (int i = 0; i < N; i++) {
		int tmp;
		scanf("%d", &tmp);
		if (i == 0) {
			a[tmp] += 1;
			cur.insert(node(tmp, a[tmp]));
		}
		else {
			printf("%d: ", tmp);
			int curK = cur.size() < K ? cur.size() : K;
			int index = 0;
			for (auto j = cur.begin(); j != cur.end(); j++) {
				index++;
				if (index != curK) {
					printf("%d ", (*j).v);
				}
				else {
					printf("%d\n", (*j).v);
					break;
				}
			}
			if (a[tmp] == 0) {
				a[tmp] += 1;
				cur.insert(node(tmp, a[tmp]));
			}
			else {
				cur.erase(node(tmp, a[tmp]));
				a[tmp] += 1;
				cur.insert(node(tmp, a[tmp]));
			}
		}

	}
	return 0;
}