输入:
    departments: dict, key是部门名, value是每个部门对应所有员工ID数组
    required_department: dict, 该任务需要参与的部门和需要的人数

    输出:
    所有可能的员工组合

    例:
    <br />({ frontend: [1, 2], backend: [3, 4],<br />devops: [5] }, { frontend: 2, backend: 1 }) => [ [1,2,3], [1,2,4]]<br />

    1. frontend = [1, 2]
    2. backend = [3, 4]
    3. devops = [5]
    6. #l为数组，n为n个元素组合
    7. def combine(l, n):
    8.     answers = []
    9.     one = [0] * n
    10.     def next_c(li = 0, ni = 0):
    11.         if ni == n:
    12.             answers.append(copy.copy(one))
    13.             return
    14.         for lj in range(li, len(l)):
    15.             one[ni] = l[lj]
    16.             next_c(lj + 1, ni + 1)
    17.     next_c()
    18.     return answers
    21. def department(require):
    22.     fr = combine(frontend, require.get("frontend", 0))
    23.     ba = combine(backend, require.get("backend"))
    24.     de = combine(devops, require.get("devops"))
    25.     ret = []
    26.     for f in fr:
    27.         for b in ba:
    28.             for d in de:
    29.                 ret.append(f + b + d)
    31.     return ret
    33. print('部门任务出席所有组合', department({"frontend": 2, "backend": 1, "devops": 0}))

    方法二

    1. # 2 部门协调
    3. frontend = [1, 2]
    4. backend = [3, 4]
    5. devops = [5]
    7. from itertools import combinations
    8. from itertools import product
    9. def department(require):
    10.     fr = list(combinations(frontend, require.get("frontend", 0)))
    11.     ba = list(combinations(backend, require.get("backend", 0)))
    12.     de = list(combinations(devops, require.get("devops", 0)))
    13.     ret = list(product(fr, ba, de))
    14.     result = []
    15.     for r in ret:
    16.         result.append(list(r[0]) + list(r[1]) + list(r[2]))
    17.     return result
    19. print('部门任务出席所有组合', department({"frontend": 2, "backend": 1, "devops": 0}))