利用遗传算法编写有多个起点多个货车的包括障碍物的VRP问题的代码
时间: 2023-12-10 19:39:11 浏览: 37
遗传算法可以用来解决VRP问题,其中包括多个起点多个货车和障碍物。下面是一个基本的遗传算法框架,可以用来解决这个问题。
1. 初始化种群:生成一组随机的解决方案,即每个货车的路径和顺序。
2. 评估适应度:根据每个解决方案的总距离和避免障碍物的程度,计算适应度。
3. 选择交叉:从种群中选择一组父代,用交叉算子将它们组合成一组新的解决方案。
4. 变异操作:对新解决方案进行变异操作,例如随机交换两个节点或移动一个节点。
5. 评估适应度:计算每个新解决方案的适应度。
6. 选择下一代:从父代和子代中选择最优的一组解决方案,作为下一代种群。
7. 重复步骤2-6,直到达到最大迭代次数或找到满意的解决方案。
下面是一个示例代码,用遗传算法解决VRP问题:
```python
import random
# 初始化参数
NUM_CARS = 3 # 货车数量
NUM_CUSTOMERS = 10 # 客户数量
MAX_ITERATIONS = 100 # 最大迭代次数
POPULATION_SIZE = 100 # 种群大小
MUTATION_RATE = 0.1 # 变异率
# 客户和货车的坐标
customers = [(random.uniform(0, 100), random.uniform(0, 100)) for i in range(NUM_CUSTOMERS)]
cars = [(random.uniform(0, 100), random.uniform(0, 100)) for i in range(NUM_CARS)]
# 障碍物坐标
obstacles = [(20, 20), (30, 30), (40, 40)]
# 计算两点之间的距离
def distance(a, b):
return ((a[0] - b[0]) ** 2 + (a[1] - b[1]) ** 2) ** 0.5
# 计算路径总长度
def get_total_distance(path):
total_distance = 0
for i in range(len(path) - 1):
total_distance += distance(customers[path[i]], customers[path[i+1]])
return total_distance
# 避免障碍物
def avoid_obstacles(path):
for o in obstacles:
if o in path:
return False
return True
# 初始化种群
population = []
for i in range(POPULATION_SIZE):
paths = []
for j in range(NUM_CARS):
path = [0] # 起点
while True:
next_customer = random.randint(1, NUM_CUSTOMERS - 1)
if next_customer not in path:
path.append(next_customer)
if len(path) >= 3 and avoid_obstacles(path):
break
paths.append(path + [0]) # 终点
population.append(paths)
# 进化过程
for i in range(MAX_ITERATIONS):
# 评估适应度
fitness = []
for j in range(len(population)):
total_distance = sum([get_total_distance(path) for path in population[j]])
fitness.append(1 / total_distance)
# 选择交叉
parents = random.choices(population, weights=fitness, k=2)
child = []
for j in range(NUM_CARS):
if random.random() < 0.5:
child.append(parents[0][j])
else:
child.append(parents[1][j])
# 变异操作
if random.random() < MUTATION_RATE:
car_index = random.randint(0, NUM_CARS - 1)
node_index1 = random.randint(1, len(child[car_index]) - 2)
node_index2 = random.randint(1, len(child[car_index]) - 2)
child[car_index][node_index1], child[car_index][node_index2] = child[car_index][node_index2], child[car_index][node_index1]
# 更新种群
population.append(child)
fitness.append(1 / sum([get_total_distance(path) for path in child]))
sorted_population = [x for _, x in sorted(zip(fitness, population), reverse=True)]
population = sorted_population[:POPULATION_SIZE]
# 输出最优解决方案
best_solution = sorted_population[0]
print(best_solution)
```