Problem description:

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:

Every worker in the paid group should be paid in the ratio of their quality compared to other workers in the paid group.
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 <= K <= N <= 10000, where N = quality.length = wage.length
1 <= quality[i] <= 10000
1 <= wage[i] <= 10000
Answers within 10^-5 of the correct answer will be considered correct.

Solution:

The idea is to use heap to sort the ratio of (wage/quality).
With the priority_queue, we can ensure that we find the higher quality but cheaper worker.
When the heap reaches the predefined size, calculate the total wages again.

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class Solution {
public:
    double mincostToHireWorkers(vector<int>& q, vector<int>& w, int K) {
        vector<vector<double>> workers;
        for (int i = 0; i < q.size(); ++i)
            workers.push_back({(double)(w[i]) / q[i], (double)q[i]});
        sort(workers.begin(), workers.end());
        
        //for(auto worker: workers)
        //    printf("%f, %f\n", worker[0], worker[1]);
        
        double res = DBL_MAX, qsum = 0;
        priority_queue<int> pq;
        //cout<<endl;
        for (auto worker: workers) {
            //printf("")
            qsum += worker[1], pq.push(worker[1]);
            if (pq.size() > K) qsum -= pq.top(), pq.pop();
            //printf("res= %f, qsum*worker[0]= %f\n", res, qsum * worker[0]);
            if (pq.size() == K) res = min(res, qsum * worker[0]);
            
        }
        return res;
    }
};

time complexity: $O(nlogn)$, $O(n)$ for the worker iteration, $O(logn)$ because heap insertion
space complexity: $O(n)$
reference:
https://www.cnblogs.com/lightwindy/p/9795918.html
https://goo.gl/qmbR5h