2023-05-18：有 n 名工人。 给定两个数组 quality 和 wage ，

【资料图】

其中，quality[i] 表示第 i 名工人的工作质量，其最低期望工资为 wage[i] 。

现在我们想雇佣 k 名工人组成一个工资组。在雇佣 一组 k 名工人时，

我们必须按照下述规则向他们支付工资：

对工资组中的每名工人，应当按其工作质量与同组其他工人的工作质量的比例来支付工资。

工资组中的每名工人至少应当得到他们的最低期望工资。

给定整数 k ，返回 组成满足上述条件的付费群体所需的最小金额。

输入： quality = [10,20,5], wage = [70,50,30], k = 2。

输出： 105.00000。

答案2023-05-18：

1.构造 Employee 结构体，存储每个员工的工作质量和垃圾系数（wage / quality）。

2.按照垃圾系数从小到大对所有员工进行排序。

3.维护一个大小为 k 的小根堆，表示当前最低期望工资组中的 k 名工人的工作质量。

4.遍历所有员工，如果堆未满，则将该员工加入堆中并更新最低期望工资。如果堆已满，则检查当前员工能否替换堆顶元素，如果可以，则弹出堆顶元素并将当前员工入堆，更新最低期望工资。

5.最终返回最低期望工资即可。

注意事项：

使用 golang 内置的 container/heap 库来实现小根堆。

在比较垃圾系数大小时，需要使用小于等于号，因为可能存在两个员工的垃圾系数完全相等的情况。

时间复杂度：排序所需的时间为 O(nlogn)，遍历员工数组时每个员工会入堆一次，出堆一次，即共进行了 2n 次操作，而小根堆的插入和弹出操作时间复杂度均为 O(logk)，因此总时间复杂度为 O(nlogn + nlogk)。

空间复杂度：除给定数组外，我们还需要构造 Employee 结构体，以及维护大小为 k 的小根堆，因此需要额外使用 O(n) 空间来存储结构体数组，并且在堆满时可能需要对堆进行调整，最多需要 O(k) 的额外空间。因此总空间复杂度为 O(n + k)。

package mainimport ("container/heap""fmt""sort")type Employee struct {RubbishDegree float64Quality       int}func NewEmployee(w, q int) Employee {return Employee{RubbishDegree: float64(w) / float64(q), Quality: q}}type EmployeeHeap []intfunc (h EmployeeHeap) Len() int            { return len(h) }func (h EmployeeHeap) Less(i, j int) bool  { return h[i] > h[j] }func (h EmployeeHeap) Swap(i, j int)       { h[i], h[j] = h[j], h[i] }func (h *EmployeeHeap) Push(x interface{}) { *h = append(*h, x.(int)) }func (h *EmployeeHeap) Pop() interface{} {n := len(*h)x := (*h)[n-1]*h = (*h)[:n-1]return x}func min(a, b float64) float64 {if a < b {return a}return b}func mincostToHireWorkers(quality []int, wage []int, k int) float64 {n := len(quality)employees := make([]Employee, n)for i := range quality {employees[i] = NewEmployee(wage[i], quality[i])}sort.Slice(employees, func(i, j int) bool {return employees[i].RubbishDegree <= employees[j].RubbishDegree})minTops := &EmployeeHeap{}heap.Init(minTops)ans, qualitySum := 1e9, 0for i := 0; i < n; i++ {curQuality := employees[i].Qualityif minTops.Len() < k { // 堆没满qualitySum += curQualityheap.Push(minTops, curQuality)if minTops.Len() == k {ans = min(ans, float64(qualitySum)*employees[i].RubbishDegree)}} else { // 来到当前员工的时候，堆是满的！// 当前员工的能力，可以把堆顶干掉，自己进来！if top := (*minTops)[0]; top > curQuality {heap.Pop(minTops)qualitySum += curQuality - topheap.Push(minTops, curQuality)ans = min(ans, float64(qualitySum)*employees[i].RubbishDegree)}}}return ans}func main() {quality := []int{10, 20, 5}wage := []int{70, 50, 30}k := 2result := mincostToHireWorkers(quality, wage, k)fmt.Println(result)}

struct Employee {    rubbish_degree: f64,    quality: i32,}impl Employee {    fn new(w: i32, q: i32) -> Self {        let rubbish_degree = w as f64 / q as f64;        Self {            rubbish_degree,            quality: q,        }    }}fn mincost_to_hire_workers(quality: Vec, wage: Vec, k: i32) -> f64 {    let n = quality.len();    let mut employees = Vec::with_capacity(n);    for i in 0..n {        employees.push(Employee::new(wage[i], quality[i]));    }    // 只根据垃圾指数排序    // 要价 / 能力    employees.sort_by(|a, b| a.rubbish_degree.partial_cmp(&b.rubbish_degree).unwrap());    // 请维持力量最小的前K个力量    // 大根堆！门槛堆！    let mut min_tops = std::collections::BinaryHeap::new();    let mut ans = std::f64::MAX;    let mut quality_sum = 0;    for i in 0..n {        // i : 依次所有员工的下标        // quality_sum : 进入堆的力量总和！        // cur_quality当前能力        let cur_quality = employees[i].quality;        if min_tops.len() < k as usize {            // 堆没满            quality_sum += cur_quality;            min_tops.push(cur_quality);            if min_tops.len() == k as usize {                ans = ans.min(quality_sum as f64 * employees[i].rubbish_degree);            }        } else {            // 来到当前员工的时候，堆是满的！            // 当前员工的能力，可以把堆顶干掉，自己进来！            if let Some(top) = min_tops.peek() {                if *top > cur_quality {                    quality_sum += cur_quality - min_tops.pop().unwrap();                    min_tops.push(cur_quality);                    ans = ans.min(quality_sum as f64 * employees[i].rubbish_degree);                }            }        }    }    ans}fn main() {    let quality = vec![10, 20, 5];    let wage = vec![70, 50, 30];    let k = 2;    let result = mincost_to_hire_workers(quality, wage, k);    println!("{}", result);}

#include #include #include typedef struct Employee {    double rubbishDegree;    int quality;} Employee;int cmp(const void* a, const void* b) {    Employee* pa = (Employee*)a;    Employee* pb = (Employee*)b;    if (pa->rubbishDegree < pb->rubbishDegree) {        return -1;    }    else if (pa->rubbishDegree > pb->rubbishDegree) {        return 1;    }    else {        return 0;    }}double mincostToHireWorkers(int* quality, int qualitySize, int* wage, int wageSize, int k) {    int n = qualitySize;    Employee* employees = (Employee*)malloc(n * sizeof(Employee));    for (int i = 0; i < n; i++) {        employees[i].quality = quality[i];        employees[i].rubbishDegree = (double)wage[i] / (double)quality[i];    }    qsort(employees, n, sizeof(Employee), cmp);    int* minTops = (int*)malloc(k * sizeof(int));    int topIndex = -1;    double ans = INT_MAX;    int qualitySum = 0;    for (int i = 0; i < n; i++) {        int curQuality = employees[i].quality;        if (topIndex < k - 1) {            qualitySum += curQuality;            minTops[++topIndex] = curQuality;            if (topIndex == k - 1) {                ans = qualitySum * employees[i].rubbishDegree;            }        }        else {            if (minTops[0] > curQuality) {                qualitySum += curQuality - minTops[0];                minTops[0] = curQuality;                ans = qualitySum * employees[i].rubbishDegree;                // 调整，使堆继续保持最小堆的性质                for (int j = 0; j < k - 1; j++) {                    if (minTops[j] > minTops[j + 1]) {                        int tmp = minTops[j];                        minTops[j] = minTops[j + 1];                        minTops[j + 1] = tmp;                    }                    else {                        break;                    }                }            }        }    }    free(employees);    free(minTops);    return ans;}int main() {    int quality[] = { 10,20,5 };    int wage[] = { 70,50,30 };    int k = 2;    double result = mincostToHireWorkers(quality, sizeof(quality) / sizeof(int), wage, sizeof(wage) / sizeof(int), k);    printf("%lf\n", result);     return 0;}

#include #include #include #include #include using namespace std;struct Employee {    double rubbishDegree;    int quality;    Employee() = default;    Employee(int w, int q) : rubbishDegree(static_cast(w) / static_cast(q)), quality(q) {}    bool operator<(const Employee& other) const {        return rubbishDegree < other.rubbishDegree;    }};double mincostToHireWorkers(vector& quality, vector& wage, int k) {    int n = quality.size();    vector employees(n);    for (int i = 0; i < n; i++) {        employees[i] = Employee(wage[i], quality[i]);    }    // 只根据垃圾指数排序    sort(employees.begin(), employees.end());    // 请维持力量最小的前K个力量    // 大根堆！门槛堆！    priority_queue minTops;    double ans = numeric_limits::max();    for (int i = 0, qualitySum = 0; i < n; i++) {        // i : 依次所有员工的下标        // qualitySum : 进入堆的力量总和！        // curQuality当前能力        int curQuality = employees[i].quality;        if (minTops.size() < k) { // 堆没满            qualitySum += curQuality;            minTops.push(curQuality);            if (minTops.size() == k) {                ans = min(ans, qualitySum * employees[i].rubbishDegree);            }        }        else { // 来到当前员工的时候，堆是满的！            // 当前员工的能力，可以把堆顶干掉，自己进来！            if (minTops.top() > curQuality) {                //              qualitySum -= minTops.top();                //              qualitySum += curQuality;                //              minTops.pop();                //              minTops.push(curQuality);                qualitySum += curQuality - minTops.top();                minTops.pop();                minTops.push(curQuality);                ans = min(ans, qualitySum * employees[i].rubbishDegree);            }        }    }    return ans;}int main() {    vector quality = { 10, 20, 5 };    vector wage = { 70, 50, 30 };    int k = 2;    double result = mincostToHireWorkers(quality, wage, k);    cout << result << endl; // 105    return 0;}

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