python – 最大化消耗能量

提供三种类型的食物,即肉,蛋糕和比萨饼
和N个不同的商店出售它,我只能从中挑选一种食物
每家商店.此外,我只能购买A,B和C号码中的物品,其中“A”表示肉类,总数为“A”个不同的商店(参见示例).我的任务是
消耗食物,这样我就可以获得最大的能量.
例,

10            <= number of stores <br>
5 3 2         <= out of 10 stores I can pick meat from 5 stores only. Similarly,
                 I can pick cake from 3 out of 10 stores...
56 44 41    1 <= Energy level of meat, cake and pizza - (56, 44, 41) for first store.<br> 
56 84 45    2
40 98 49    3
91 59 73    4
69 94 42    5
81 64 80    6
55 76 26    7
63 24 22    8
81 60 44    9
52 95 11    10

所以为了最大化我的能量,我可以消耗…

>商店数量的肉：

[1, 4, 7, 8, 9] => [56, 91, 55, 63, 81]

>商店号码的蛋糕：

[3, 5, 10] => [98, 94, 95]

>商店数量的比萨饼：

[2, 6] => [45, 80]

这导致我最终获得758的最大能量水平.

由于我是动态编程的新手,我试图通过生成独特的组合来解决它,比如

10C5 *(10-5)C3 *(10-5-3)C2 = 2520

这是我的代码,

nStores = 10
a, b, c = 5, 3, 2
matrix = [
    [56,44,41],
    [56,84,45],
    [40,98,49],
    [91,59,73],
    [69,94,42],
    [81,64,80],
    [55,76,26],
    [63,24,22],
    [81,60,44],
    [52,95,11]
]

count = a + b + c
data = []
allOverCount = [i for i in range(count)]
def genCombination(offset, depth, passedData, reductionLevel = 3):
    if (depth == 0):
        first = set(data)
        if reductionLevel ==  3:
            return genCombination(0,b,[i for i in allOverCount if i not in first], reductionLevel=2)
        elif reductionLevel ==  2:
            return genCombination(0,c,[i for i in allOverCount if i not in first], reductionLevel=1)
        elif reductionLevel == 1:
            xAns = 0
            for i in range(len(data)):
                if i < a:
                    xAns += matrix[data[i]][0]
                elif i < a + b:
                    xAns += matrix[data[i]][1]
                else:
                    xAns += matrix[data[i]][2]
            return xAns
    oneData = 0
    for i in range(offset, len(passedData) - depth + 1 ):
        data.append(passedData[i])
        oneData = max(oneData, genCombination(i+1, depth-1, passedData, reductionLevel))
        del data[-1]
    return oneData
passedData = [i for i in range(count)]
finalOutput = genCombination(0,a,passedData)
print(finalOutput)

我知道这不是正确的做法.我该如何优化它？

解决方法:

这是使用纸浆线性规划(https://pypi.org/project/PuLP)的解决方案,它为我提供了最佳解决方案

Maximum energy level: 758.0
Mapping of stores per foodtype: {1: [9, 2, 4], 0: [3, 8, 0, 6, 7], 2: [1, 5]}

我认为性能应该优于手动编码的穷举求解器.

from collections import defaultdict

import pulp

# data
nStores = 10
a, b, c = max_stores = 5, 3, 2
matrix = [
    [56, 44, 41],
    [56, 84, 45],
    [40, 98, 49],
    [91, 59, 73],
    [69, 94, 42],
    [81, 64, 80],
    [55, 76, 26],
    [63, 24, 22],
    [81, 60, 44],
    [52, 95, 11]
]

# create an LP problem
lp = pulp.LpProblem("maximize energy", sense=pulp.LpMaximize)

# create the list of indices for the variables
# the variables are binary variables for each combination of store and food_type
# the variable alpha[(store, food_typeà] = 1 if the food_type is taken from the store
index = {(store, food_type) for store in range(nStores) for food_type in range(3)}
alpha = pulp.LpVariable.dicts("alpha", index, lowBound=0, cat="Binary")

# add the constrain on max stores
for food_type, n_store_food_type in enumerate(max_stores):
    lp += sum(alpha[(store, food_type)] for store in range(nStores)) <= n_store_food_type

# only one food type can be taken per store
for store in range(nStores):
    lp += sum(alpha[(store, food_type)] for food_type in range(3)) <= 1

# add the objective to maximise
lp += sum(alpha[(store, food_type)] * matrix[store][food_type] for store, food_type in index)

# solve the problem
lp.solve()

# collect the results
stores_for_foodtype = defaultdict(list)
for (store, food_type) in index:
    # check if the variable is active
    if alpha[(store, food_type)].varValue:
        stores_for_foodtype[food_type].append(store)

print(f"Maximum energy level: {lp.objective.value()}")
print(f"Mapping of stores per foodtype: {dict(stores_for_foodtype)}")

标签：python,algorithm,dynamic-programming
来源： https://codeday.me/bug/20190929/1829897.html