leetcode 857

问题

There are N workers. The i-th worker has a quality[i] and a minimum wage expectation wage[i].

Now we want to hire exactly K workers to form a paid group. When hiring a group of K workers, we must pay them according to the following rules:

  1. Every worker in the paid group should be paid in the ratio of their quality compared to other workers in the paid group.
  2. Every worker in the paid group must be paid at least their minimum wage expectation.

Return the least amount of money needed to form a paid group satisfying the above conditions.

Example 1:

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Input: quality = [10,20,5], wage = [70,50,30], K = 2
Output: 105.00000
Explanation: We pay 70 to 0-th worker and 35 to 2-th worker.

Example 2:

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Input: quality = [3,1,10,10,1], wage = [4,8,2,2,7], K = 3
Output: 30.66667
Explanation: We pay 4 to 0-th worker, 13.33333 to 2-th and 3-th workers seperately.

Note:

  1. 1 <= K <= N <= 10000, where N = quality.length = wage.length
  2. 1 <= quality[i] <= 10000
  3. 1 <= wage[i] <= 10000
  4. Answers within 10^-5 of the correct answer will be considered correct.

分析

首先考虑，假设我们现在已经有了K个工人，那么一共应该给他们发多少工资呢？假设每个工人单位工作量所需要的工资为（quality[i]\wage[i]），那么k个工人所需要支付的总的工资数就是
$$
\sum \limits_{i} quality[i]* \min \limits_{i} \{ quality[i] \div wage[i]\}
$$

那么我们可以按照每个工人的每单位工作量所需要的工资数（quality[i]\wage[i]） 对工人进行从小到大的排序。然后依次尝试将每个工人加入到队伍中时，工人所需要支付的总的工资数的情况。若当前队伍人数已经大于等于K时，则从员队伍中取出quality最高的员工，再尝试将新员工加入到队伍中去。

代码

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class Solution {
public:
    const double eps = 10e-6;
    double mincostToHireWorkers(vector<int>& quality, vector<int>& wage, int K) {
        vector<pair<int, double>> workers;
        for (int i = 0; i < quality.size(); ++i) {
            workers.push_back({quality[i], (double)wage[i]/quality[i]});
        }
        
        sort(workers.begin(), workers.end(), [](
            const pair<int,double>&a, const pair<int,double> &b) {
            return a.second < b.second;});
        
        priority_queue<int> pq;
        int sumQuality = 0;
        double ans = 0.0;
        for (auto worker: workers) {
            if (pq.size() == K) {
                int x = pq.top();
                pq.pop();
                sumQuality -= x;
            }
            pq.push(worker.first);
            sumQuality += worker.first;
            
            if (pq.size() == K) {
                double curPaid = sumQuality*worker.second;
                if (ans <= eps && ans >= -eps)
                    ans = curPaid;
                else
                    ans = min(ans, curPaid);
            }
        }
        return ans;
    }
};