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Java 排班算法解决方案
对于这个排班算法,我们可以将其分为以下几个步骤:
- 数据初始化
- 接收入参:排班月出勤人数,上班天数
- 初始化每个员工的出勤时间
- 初始化每个班次的出勤人数占当月出勤人数的百分比
- 建立排班规则
- 根据要求5和6,确定每个员工每次上大夜班需要连续排两天,大夜后需接2天休息
- 根据要求4,小夜、小夜2不能接常班、常1班次
- 根据要求7,确定每天各个班次出勤人数占当月出勤人数的百分比
- 排班过程
- 遍历每个员工
- 对于每个员工,遍历每天的出勤情况
- 如果员工出勤,则根据排班规则确定员工上哪个班次
- 如果员工不出勤,则将其视为休息
- 输出结果
- 将每个员工的出勤情况存储在一个List<map>中,每个map代表这个人当月1-30号所上的班次</map>
以下是一个简单的 Java 实现:
import java.util.*;
class Employee {
String name;
int attendanceDays;
List<Integer> attendanceDates;
Employee(String name, int attendanceDays, List<Integer> attendanceDates) {
this.name = name;
this.attendanceDays = attendanceDays;
this.attendanceDates = attendanceDates;
}
}
class Shift {
String name;
int attendanceCount;
Shift(String name, int attendanceCount) {
this.name = name;
this.attendanceCount = attendanceCount;
}
}
public class Scheduler {
public List<Map<String, String>> schedule(List<Employee> employees, int totalAttendanceDays, Map<String, Shift> shiftMap) {
List<Map<String, String>> result = new ArrayList<>();
for (Employee employee : employees) {
Map<String, String> employeeSchedule = new HashMap<>();
for (int i = 1; i <= totalAttendanceDays; i++) {
if (employee.attendanceDates.contains(i)) {
Shift shift = getShift(employee, i, shiftMap);
employeeSchedule.put(String.valueOf(i), shift.name);
} else {
employeeSchedule.put(String.valueOf(i), "休息");
}
}
result.add(employeeSchedule);
}
return result;
}
private Shift getShift(Employee employee, int date, Map<String, Shift> shiftMap) {
Shift shift = null;
if (date % 3 == 0) {
shift = shiftMap.get("小夜");
} else if (date % 3 == 1) {
shift = shiftMap.get("小夜2");
} else if (date % 3 == 2) {
shift = shiftMap.get("常班");
} else {
shift = shiftMap.get("大夜");
}
return shift;
}
public static void main(String[] args) {
List<Employee> employees = new ArrayList<>();
employees.add(new Employee("员工1", 30, Arrays.asList(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30)));
employees.add(new Employee("员工2", 30, Arrays.asList(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30)));
employees.add(new Employee("员工3", 30, Arrays.asList(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30)));
Map<String, Shift> shiftMap = new HashMap<>();
shiftMap.put("常班", new Shift("常班", 50));
shiftMap.put("小夜", new Shift("小夜", 20));
shiftMap.put("小夜2", new Shift("小夜2", 20));
shiftMap.put("大夜", new Shift("大夜", 30));
Scheduler scheduler = new Scheduler();
List<Map<String, String>> result = scheduler.schedule(employees, 30, shiftMap);
for (Map<String, String> employeeSchedule : result) {
System.out.println(employeeSchedule);
}
}
}
请注意,这只是一个简单的实现,实际上可能需要更多的逻辑和调整来满足具体的需求。