Minimum Cost to Hire K Workers LeetCode solution in C++

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class Solution {
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 public:
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  double mincostToHireWorkers(vector<int> &quality, vector<int> &wage, int k) {
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    priority_queue<pair<double, int>, vector<pair<double, int>>, greater<pair<double, int>>> pqmn;
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    priority_queue<int> pqmx;
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    int n = quality.size();
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    for (int i = 0; i < n; i++) {
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      pqmn.push({(double)wage[i] / quality[i], quality[i]});
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    }
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    double SumQuality = 0;
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    double ans = INT_MAX;
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    while (!pqmn.empty()) {
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      pair<double, int> p = pqmn.top();
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      SumQuality += p.second;
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      pqmx.push(p.second);
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      pqmn.pop();
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      if (pqmx.size() > k) {
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        SumQuality -= pqmx.top();
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        pqmx.pop();
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      }
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      if (pqmx.size() == k) {
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        ans = min(ans, SumQuality * p.first);
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      }
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    }
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    return ans;
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  }
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};