提供了三种类型的食物，即肉，蛋糕和比萨饼，在N个不同的商店出售 食物 ， 我只能从每个商店中选择一种食物 。另外，我只能以A，B和C编号购买商品，其中“ A”表示从不同商店的“ A”总数中购买肉类（请参见示例）。我的任务是食用食物，以使我拥有最大的能量。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
因此，要最大限度地利用能量，我可以消耗…
商店编号中的肉类：
[1, 4, 7, 8, 9] => [56, 91, 55, 63, 81]
店铺编号的蛋糕：
[3, 5, 10] => [98, 94, 95]
商店编号的披萨：
[2, 6] => [45, 80]
这使我最终获得758的最大能量。
由于我是动态编程的新手，因此我尝试通过生成独特的组合（例如，
10 C 5 * （10-5） C 3 * （10-5-3） C 2 = 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方法。我该如何优化？
这是使用线性编程通过纸浆（https://pypi.org/project/PuLP）的解决方案，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]}
我认为该性能应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)}")
