计算调度问题中的雇佣天数

Question

在调度问题中，我还想尽量减少总雇佣天数。

如果员工在某天之前和之后工作，他/她就会在某一天被雇用。

这是一个小的工作示例：

import random
from ortools.sat.python import cp_model

model = cp_model.CpModel()
solver = cp_model.CpSolver()

employees = range(3)
days = range(10)

works_day = {(e, d): model.NewBoolVar(f'{e}_works_{d}')
             for e in employees for d in days}
hired_day = {(e, d): model.NewBoolVar(f'{e}_employed_{d}')
             for e in employees for d in days}

# random example
for boolean in works_day.values():
    model.Add(boolean == random.choice([0, 1]))

# give value to hired_day
add_hired_days()

# solve
print('Variables:', len(model.Proto().variables))
print('Constraints:', len(model.Proto().constraints))
status = solver.Solve(model)

for e in employees:
    print()
    print('Employee', e)
    for d in days:
        print('Works', solver.Value(works_day[e, d]),
              'Hired', solver.Value(hired_day[e, d]))

add_hired_days 在哪里：

def add_hired_days():
    for idx, d in enumerate(days):
        for e in employees:
            model.AddImplication(works_day[e, d], hired_day[e, d])
            previous = [works_day[e, d] for d in days[:idx + 1]]
            following = [works_day[e, d] for d in days[idx:]]

            # too many variables
            works_previous = model.NewBoolVar('')
            works_following = model.NewBoolVar('')

            model.AddBoolOr(previous).OnlyEnforceIf(works_previous)
            model.AddBoolAnd([d.Not() for d in previous
                              ]).OnlyEnforceIf(works_previous.Not())

            model.AddBoolOr(following).OnlyEnforceIf(works_following)
            model.AddBoolAnd([d.Not() for d in following
                              ]).OnlyEnforceIf(works_following.Not())

            model.AddBoolAnd([works_previous, works_following
                              ]).OnlyEnforceIf(hired_day[e, d])
            model.AddBoolOr([works_previous.Not(),
                             works_following.Not()
                             ]).OnlyEnforceIf(hired_day[e, d].Not())

有没有办法在不创建这么多变量和约束的情况下做到这一点？

Answer 1

如果一个员工一个月工作 n 天，他需要被雇用 n - 2 次。

Answer 2

按照Laurent的建议解决。

import random
from ortools.sat.python import cp_model

if __name__ == '__main__':
    model = cp_model.CpModel()
    solver = cp_model.CpSolver()

    employees = range(3)
    days = range(10)
    horizon = len(days) - 1

    works_day = {(e, d): model.NewBoolVar(f'{e}_works_{d}')
                 for e in employees for d in days}
    hired_days = [
        model.NewIntVar(0, len(days), f'{e}_hired') for e in employees
    ]
    first_day = [
        model.NewIntVar(0, horizon, f'{e}_first_day') for e in employees
    ]
    last_day = [
        model.NewIntVar(0, horizon, f'{e}_last_day') for e in employees
    ]

    # random example
    for boolean in works_day.values():
        model.Add(boolean == random.choice([0, 1]))

    for e in employees:
        v1 = [model.NewIntVar(0, horizon, '') for _ in days]
        v2 = [model.NewIntVar(0, horizon, '') for _ in days]
        for d in days:
            model.Add(v1[d] == d * works_day[e, d])
            model.Add(v2[d] == horizon + works_day[e, d] * (d - horizon))
        model.AddMinEquality(first_day[e], v2)
        model.AddMaxEquality(last_day[e], v1)
        model.Add(hired_days[e] == last_day[e] - first_day[e] + 1)

    # solve
    status = solver.Solve(model)

    for e in employees:
        print()
        print('Employee', e)
        for d in days:
            print('Works', solver.Value(works_day[e, d]))
        print('First day:', solver.Value(first_day[e]), 'Last day:',
              solver.Value(last_day[e]), 'Hired:', solver.Value(hired_days[e]))