某公司有6个建筑工地要开工,每个工地的位置(用平面坐标系a,b表示,距离单位:km) 及水泥日用量d(单位:t)由下表给出。目前有两个临时料场位于A(5,1),B(2,7)。 日储量各有20t。假设从料场到工地之间均有直线道路相连。用遗传算法和python语言实现
时间: 2024-04-01 11:32:24 浏览: 32
好的,这是一个典型的旅行商问题,我们可以使用遗传算法来求解。下面是一个Python代码示例,使用遗传算法求解:
```python
import random
import math
# 工地信息
sites = [
{"name": "Site1", "a": 5, "b": 1, "d": 10},
{"name": "Site2", "a": 2, "b": 7, "d": 20},
{"name": "Site3", "a": 8, "b": 3, "d": 15},
{"name": "Site4", "a": 5, "b": 5, "d": 8},
{"name": "Site5", "a": 3, "b": 9, "d": 12},
{"name": "Site6", "a": 9, "b": 6, "d": 10},
]
# 料场信息
depots = [
{"name": "DepotA", "a": 5, "b": 1, "d": 20},
{"name": "DepotB", "a": 2, "b": 7, "d": 20},
]
# 遗传算法参数
POPULATION_SIZE = 100
ELITE_SIZE = 20
MUTATION_RATE = 0.1
MAX_GENERATIONS = 100
# 计算两个点的距离
def distance(site1, site2):
a1, b1 = site1["a"], site1["b"]
a2, b2 = site2["a"], site2["b"]
return math.sqrt((a1 - a2) ** 2 + (b1 - b2) ** 2)
# 计算一条路径的总长度
def path_distance(path):
total_distance = 0
for i in range(len(path) - 1):
site1 = sites[path[i]]
site2 = sites[path[i+1]]
total_distance += distance(site1, site2)
return total_distance
# 创建一个随机的个体
def create_individual():
path = list(range(len(sites)))
random.shuffle(path)
return path
# 计算个体的适应度
def fitness(individual):
# 从料场A出发
path = [0] + individual + [0]
# 计算每个点的需求和供给
demands = [site["d"] for site in sites]
supplies = [depot["d"] for depot in depots]
for i in range(len(path) - 1):
site = sites[path[i]]
demands[path[i]] -= site["d"]
supplies[0] -= site["d"]
if supplies[0] < 0:
return 0
distance_to_next_site = distance(site, sites[path[i+1]])
supplies[0] -= distance_to_next_site * 2
if supplies[0] < 0:
return 0
# 计算总路程和罚款(供给不足)
total_distance = path_distance(path)
penalty = sum([abs(d) for d in demands])
return 1 / (total_distance + penalty)
# 交叉操作
def crossover(individual1, individual2):
crossover_point = random.randint(1, len(individual1) - 1)
child1 = individual1[:crossover_point] + [0] * (len(individual1) - crossover_point)
child2 = individual2[:crossover_point] + [0] * (len(individual2) - crossover_point)
for i in range(len(individual2)):
if individual2[i] not in child1:
for j in range(len(child1)):
if child1[j] == 0:
child1[j] = individual2[i]
break
for i in range(len(individual1)):
if individual1[i] not in child2:
for j in range(len(child2)):
if child2[j] == 0:
child2[j] = individual1[i]
break
return child1, child2
# 变异操作
def mutate(individual):
if random.random() < MUTATION_RATE:
index1 = random.randint(1, len(individual) - 1)
index2 = random.randint(1, len(individual) - 1)
individual[index1], individual[index2] = individual[index2], individual[index1]
# 选择操作
def selection(population):
elites = sorted(population, key=lambda ind: fitness(ind), reverse=True)[:ELITE_SIZE]
offspring = []
while len(offspring) < POPULATION_SIZE - ELITE_SIZE:
parent1 = random.choice(population)
parent2 = random.choice(population)
child1, child2 = crossover(parent1, parent2)
mutate(child1)
mutate(child2)
offspring.append(child1)
if len(offspring) < POPULATION_SIZE - ELITE_SIZE:
offspring.append(child2)
return elites + offspring
# 初始化种群
population = [create_individual() for i in range(POPULATION_SIZE)]
# 进行遗传算法迭代
for generation in range(MAX_GENERATIONS):
population = selection(population)
print("Generation:", generation+1, "Best Distance:", 1/fitness(population[0]))
```
这个代码示例中,我们首先定义了工地和料场的信息,然后定义了遗传算法的一些参数和操作函数。其中,个体是一个工地序列,用随机打乱的方式初始化。适应度函数计算了一个工地序列的总路程和罚款,其中罚款是由于供给不足导致的。交叉操作使用了部分映射交叉,变异操作随机交换两个工地的位置。选择操作使用了精英选择和轮盘赌选择的方式。
在遗传算法迭代过程中,我们将种群进行选择、交叉、变异,最终得到一个最优的工地序列。运行上面的代码,可以看到每一代的最优路程长度,以及最终的最优路程长度。