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package 第03期.mca_02;
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import java.util.Arrays;
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import java.util.PriorityQueue;
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// 有 n 名工人。 给定两个数组 quality 和 wage ，
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// 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] 。
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// 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，
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// 我们必须按照下述规则向他们支付工资：
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// 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。
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// 工资组中的每名工人至少应当得到他们的最低期望工资。
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// 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
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// 测试链接 : https://leetcode.cn/problems/minimum-cost-to-hire-k-workers/
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public class Code03_MinimumCostToHireKWorkers {
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public static class Employee {
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public double rubbishDegree;
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public int quality;
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public Employee(int w, int q) {
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rubbishDegree = (double) w / (double) q;
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quality = q;
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}
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}
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public double mincostToHireWorkers(int[] quality, int[] wage, int k) {
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int n = quality.length;
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Employee[] employees = new Employee[n];
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for (int i = 0; i < n; i++) {
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employees[i] = new Employee(wage[i], quality[i]);
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}
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Arrays.sort(employees, (a, b) -> a.rubbishDegree <= b.rubbishDegree ? -1 : 1);
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PriorityQueue<Integer> minTops = new PriorityQueue<Integer>((a, b) -> b - a);
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double ans = Double.MAX_VALUE;
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for (int i = 0, qualitySum = 0; i < n; i++) {
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int curQuality = employees[i].quality;
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if (minTops.size() < k) { // 堆没满
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qualitySum += curQuality;
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minTops.add(curQuality);
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if (minTops.size() == k) {
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ans = Math.min(ans, qualitySum * employees[i].rubbishDegree);
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}
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} else {
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if (minTops.peek() > curQuality) {
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qualitySum += curQuality - minTops.poll();
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minTops.add(curQuality);
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ans = Math.min(ans, qualitySum * employees[i].rubbishDegree);
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}
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}
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}
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return ans;
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}
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}
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