利用模拟退火算法解决时间窗车辆路径问题的代码

以下是一个简单的利用模拟退火算法求解时间窗车辆路径问题的 Python 代码：

import math
import random

# 车辆路径问题的节点类
class Node:
    def __init__(self, x, y, demand, ready_time, due_time, service_time):
        self.x = x
        self.y = y
        self.demand = demand
        self.ready_time = ready_time
        self.due_time = due_time
        self.service_time = service_time

# 计算两个节点之间的距离
def calc_distance(node1, node2):
    dx = node1.x - node2.x
    dy = node1.y - node2.y
    return math.sqrt(dx*dx + dy*dy)

# 计算路径长度
def calc_path_length(path, nodes):
    length = 0
    cur_time = 0
    cur_capacity = 0
    for i in range(len(path)):
        node = nodes[path[i]]
        if cur_capacity + node.demand > capacity:
            length += calc_distance(nodes[path[i-1]], nodes[0])
            cur_time = max(cur_time + node.service_time, node.ready_time)
            cur_capacity = node.demand
        else:
            cur_time = max(cur_time + calc_distance(nodes[path[i-1]], node) + node.service_time, node.ready_time)
            cur_capacity += node.demand
        if cur_time > node.due_time:
            return float('inf')
        if i == len(path) - 1:
            length += calc_distance(nodes[path[i]], nodes[0]])
    return length

# 模拟退火算法求解路径
def solve(nodes):
    # 初始温度、降温系数、迭代次数
    T = 100
    alpha = 0.99
    iterations = 10000
    
    # 初始解
    cur_path = list(range(1, len(nodes)))
    random.shuffle(cur_path)
    
    # 计算初始解的路径长度
    cur_length = calc_path_length(cur_path, nodes)
    
    # 退火过程
    while T > 1e-8:
        for i in range(iterations):
            # 随机交换两个节点
            x, y = random.sample(range(len(cur_path)), 2)
            new_path = cur_path[:]
            new_path[x], new_path[y] = new_path[y], new_path[x]
            
            # 计算新解的路径长度
            new_length = calc_path_length(new_path, nodes)
            
            # 根据概率接受新解或者保留当前解
            diff = new_length - cur_length
            if diff < 0 or random.random() < math.exp(-diff/T):
                cur_path = new_path
                cur_length = new_length
        
        # 降温
        T *= alpha
    
    return cur_path

# 测试代码
capacity = 10
nodes = [Node(0, 0, 0, 0, float('inf'), 0)]
nodes.append(Node(1, 1, 3, 3, 10, 2))
nodes.append(Node(2, 2, 5, 4, 10, 3))
nodes.append(Node(3, 3, 2, 5, 10, 1))
nodes.append(Node(4, 4, 4, 1, 10, 2))
nodes.append(Node(5, 5, 6, 2, 10, 3))
nodes.append(Node(6, 6, 1, 3, 10, 1))

path = solve(nodes)
print(path)
print(calc_path_length(path, nodes))

需要注意的是，这只是一个简单的实现，实际应用中需要更加细致地考虑时间窗、容量等限制条件，并进行一些优化。