linear_sum_assignment方法不支持约束或自定义目标，因此我认为这是不可能的。
然而，您可以将问题表述为混合整数线性规划问题（MILP），并通过PuLP 1来求解。为了平均分配每个工人的总成本，您可以使每个工人的最大总成本与最小总成本之间的差异最小化。以下是一个可能的表述：

Sets:

- workers = ["a", "b", "c"]
- tasks   = [1, 2, ..., 12]

Variables:

- x[w,t] = 1 iff worker w is assigned to task t, 0 otherwise
- min_val
- max_val

Model:

min  max_val - min_val

s.t. 

# each worker is assigned to exactly n_tasks_per_worker tasks

sum(x[w,t] for t in tasks) == n_tasks_per_worker for all w in workers

# each task can only be assigned once

sum(x[w,t] for w in workers) == 1 for all t in tasks

# evenly distribute the tasks

sum(x[w,t] for t in tasks) <= max_val for all w in workers
sum(x[w,t] for t in tasks) >= min_val for all w in workers

代码很简单：

import pulp
import numpy as np

workers = ["a", "b", "c"]
n_workers = len(workers)
n_tasks_per_worker = 4
n_tasks = n_workers * n_tasks_per_worker

c = np.array([[27, 42, 65, 33, 67, 45, 60, 76,  6,  6, 43, 26],
              [27, 42, 65, 33, 67, 45, 60, 76,  6,  6, 43, 26],
              [27, 42, 65, 33, 67, 45, 60, 76,  6,  6, 43, 26]])

# create the model

mdl = pulp.LpProblem("even_assignment")

# decision variables

x = {}
for w in workers:
    for t in range(n_tasks):
        x[w, t] = pulp.LpVariable(f"x[{w}, {t}]", cat="Binary")

max_val = pulp.LpVariable("max_val", cat="Continuous")
min_val = pulp.LpVariable("min_val", cat="Continuous")

# objective: minimize the difference between the maximum and the minimum

# costs per worker

mdl.setObjective(max_val - min_val)

# constraint: each worker gets assigned exactly n_tasks_per_worker

for w in workers:
    mdl.addConstraint(sum(x[w, task] for task in range(n_tasks)) == n_tasks_per_worker)

# constraint: each task can only be assigned once

for task in range(n_tasks):
    mdl.addConstraint(sum(x[w, task] for w in workers) == 1)

# constraint: evenly distribute the tasks

for i_w, w in enumerate(workers):
    assignment_cost = sum(x[w,task]*c[i_w,task] for task in range(n_tasks))
    mdl.addConstraint(assignment_cost <= max_val)
    mdl.addConstraint(assignment_cost >= min_val)

# solve the problem

mdl.solve()

# Output

for i_w, w in enumerate(workers):
    worker_cost = sum(x[w, t].varValue*c[i_w, t] for t in range(n_tasks))
    print(f"costs for worker {w}: {worker_cost:.2f}")

这给了我

costs for worker a: 165.00
costs for worker b: 167.00
costs for worker c: 164.00

[1]确切地说，PuLP不是一个解算器，它只是一个建模框架，可以将MILP传递给MILP解算器，如CBC、SCIP、HiGHS等。