Question

问题陈述：我有150个附加权重和值的对象。对象的权重可能会根据选择的顺序而改变，通常会选择约70-80个项目。我只能选择最大权重，因此一旦我找到了具有该序列的子解决方案，就需要跳过以相同序列开头的所有排列。目标是最大化价值。

我可以通过以下方式轻松创建所有排列：

from itertools import permutations
for i in permutations(list(range(150))):
    # do something with i

然而，这将创建许多我不需要检查的序列。我也可以用r来限制置换长度

from itertools import permutations
for i in permutations(list(range(150)), r=80):
    # do something with i

然而，对于非常糟糕的序列，仍然会有大量的冗余检查。此外，这可能会在“最佳”解决方案之前停止。

我可以做类似

的事情

from itertools import permutations
v = []
for i in permutations(list(range(150)), r=80):
    if v and v == i[:len(v)]:
        continue
    # do something with i
    v = i # would be some optimal subset of i

然而，由于Python仍在生成和检查序列，因此仍需要很长时间才能运行。有关如何处理此问题的任何想法？理想情况下，我可以并行运行检查。

更多背景：我正在尝试为名为Black Desert Online的游戏创建优化的资源图表（图片示例在somethinglovely.net/bdo/）。该图具有~150个资源节点，每个节点可以连接到14个根节点的子集。图上的中间节点具有与它们相关联的权重。每个城市都有可以连接的最大节点数，并且将资源节点连接到城市还有额外的重量成本。我没有成功地使用随机图生成与遗传算法相结合来“找到”最优解。此外，只是做出贪婪的选择会导致次优解决方案。我目前难以理解如何生成一个在合理的时间段内运行的强力+综合解决方案（合理的在一天内在合理的台式计算机上运行。

Answer 1

一次浏览库存清单，一个项目，并尝试打包有和没有该项目（两次递归）。当我们达到两点之一时报告解决方案：

不再需要考虑的项目
剩下的清单符合我们的体重预算。

通过积极的施工来处理剔除。

代码：

items = [
    # Description, weight
    ("petrol", 10),
    ("clothes", 8),
    ("tents", 7),
    ("beer", 16),
    ("food", 20),
    ("teddy bear", 3),
    ("tank", 25),
    ("skin lotion", 2),
    ("library", 17),
    ("mortar", 9),
    ("cut lumber", 12),
    ("sports gear", 14),
]

limit = 20

def load(inventory, max_weight, current):

    still_okay = [item for item in inventory if item[1] <= max_weight]
    if len(still_okay) == 0:
        # Can't add any more; emit solution and back up
        print "RESULT", current
    else:
        # If the rest of the list fits in our weight budget,
        #   take everything.
        if sum([item[1] for item in still_okay]) <= max_weight:
            print "RESULT", current + still_okay
        else:
            item = still_okay.pop()
            # recur on two branches: one each with and without this item
            load(still_okay, max_weight - item[1], current + [item])
            load(still_okay, max_weight, current)

load(items, limit, [])

输出：

RESULT [('sports gear', 14), ('teddy bear', 3), ('skin lotion', 2)]
RESULT [('cut lumber', 12), ('skin lotion', 2), ('teddy bear', 3)]
RESULT [('cut lumber', 12), ('teddy bear', 3)]
RESULT [('cut lumber', 12), ('tents', 7)]
RESULT [('cut lumber', 12), ('clothes', 8)]
RESULT [('mortar', 9), ('skin lotion', 2), ('teddy bear', 3)]
RESULT [('mortar', 9), ('skin lotion', 2), ('tents', 7)]
RESULT [('mortar', 9), ('skin lotion', 2), ('clothes', 8)]
RESULT [('mortar', 9), ('teddy bear', 3), ('tents', 7)]
RESULT [('mortar', 9), ('teddy bear', 3), ('clothes', 8)]
RESULT [('mortar', 9), ('tents', 7)]
RESULT [('mortar', 9), ('clothes', 8)]
RESULT [('mortar', 9), ('petrol', 10)]
RESULT [('library', 17), ('skin lotion', 2)]
RESULT [('library', 17), ('teddy bear', 3)]
RESULT [('skin lotion', 2), ('teddy bear', 3), ('tents', 7), ('clothes', 8)]
RESULT [('skin lotion', 2), ('teddy bear', 3), ('clothes', 8)]
RESULT [('skin lotion', 2), ('teddy bear', 3), ('petrol', 10)]
RESULT [('skin lotion', 2), ('beer', 16)]
RESULT [('skin lotion', 2), ('tents', 7), ('clothes', 8)]
RESULT [('skin lotion', 2), ('tents', 7), ('petrol', 10)]
RESULT [('skin lotion', 2), ('petrol', 10), ('clothes', 8)]
RESULT [('teddy bear', 3), ('beer', 16)]
RESULT [('teddy bear', 3), ('tents', 7), ('clothes', 8)]
RESULT [('teddy bear', 3), ('tents', 7), ('petrol', 10)]
RESULT [('teddy bear', 3), ('clothes', 8)]
RESULT [('teddy bear', 3), ('petrol', 10)]
RESULT [('food', 20)]
RESULT [('beer', 16)]
RESULT [('tents', 7), ('clothes', 8)]
RESULT [('tents', 7), ('petrol', 10)]
RESULT [('petrol', 10), ('clothes', 8)]

如何在剔除坏序列时生成排列？

1 个答案: