while len(own_pickup_node) > 0: route = [depot.depot_id] for i in range(min(len(own_pickup_node), 2)): minIndex = np.argmin([model.distance_matrix[depot.depot_id, own_pickup_node[j]] for j in range(0, len(own_pickup_node))]) minnode = own_pickup_node[minIndex] route.append(minnode) arrival = model.time_matrix[depot.depot_id, minnode] departure = arrival + service_time route.append(own_delivery_node[minIndex]) arrival = departure + model.time_matrix[minnode, own_delivery_node[minIndex]] departure += arrival + service_time last_node = own_delivery_node[minIndex] own_pickup_node.remove(minnode) own_delivery_node.remove(own_delivery_node[minIndex]) for j in own_pickup_node: next_minIndex = np.argmin([model.distance_matrix[last_node, j]]) next_minnode = own_pickup_node[next_minIndex] arrival = departure + model.time_matrix[last_node, next_minnode] if arrival <= model.demand_dict[next_minnode].end_time and arrival <= depot.dend_time: route.append(next_minnode) departure = arrival + service_time route.append(own_delivery_node[next_minIndex]) arrival = departure + model.time_matrix[next_minnode, own_delivery_node[next_minIndex]] departure += arrival + service_time last_node = own_delivery_node[next_minIndex] own_pickup_node.remove(next_minnode) own_delivery_node.remove(own_delivery_node[next_minIndex]) else: continue route.append(depot.depot_id) sol.route_list.append(route) vehicle_number = vehicle_number - 1 if vehicle_number == 0: break

这段代码实现了一个基于贪心算法的送货路径规划。具体来说，它会先将起点设为仓库，然后在所有尚未配对的取货点和配送点中，选取距离仓库最近的两个取货点进行配对，将它们的编号加入路径中。然后，它会根据这两个取货点所对应的配送点的时间窗口，计算出到达和离开的时间，并将这两个配送点的编号加入路径中。接着，它会在剩下的取货点和配送点中，选取距离上一个配送点最近的取货点进行配对，然后重复之前的操作，不断将取货点和配送点加入路径中，直到所有货物都配送完毕。

需要注意的是，这段代码中的距离矩阵、时间矩阵、时间窗口等信息都是由 model 对象提供的，而且这里采用了 NumPy 库中的 argmin 函数来查找距离最近的取货点。另外，这段代码中的 sol 对象则是一个 Solution 类的实例，它用于保存所有的路径方案。

相关问题

def generateOwnCarRoute(service_time, model, sol): pickup_node = copy.deepcopy(model.demand_id_list[0: 16]) own_pickup_node = [] own_delivery_node = [] route = [] sol.route_list = [] depot = model.depot_dict['d1'] vehicle_number = depot.depot_capacity departure = 0 arrival = 0 for i in pickup_node: if i not in model.crowd_pickup_node: own_pickup_node.append(i) own_delivery_node.append(i+16) while vehicle_number > 0 and len(own_pickup_node) > 0: route.append(depot.depot_id) minIndex = np.argmin([model.distance_matrix[depot.depot_id, own_pickup_node[i]] for i in range(0, len(own_pickup_node))]) minnode = own_pickup_node[minIndex] route.append(minnode) arrival = model.time_matrix[depot.depot_id, minnode] departure = arrival + service_time route.append(own_delivery_node[minIndex]) arrival = departure + model.time_matrix[minnode, own_delivery_node[minIndex]] departure += arrival + service_time last_node = own_delivery_node[minIndex] own_pickup_node.remove(minnode) own_delivery_node.remove(own_delivery_node[minIndex]) for j in own_pickup_node: next_minIndex = np.argmin([model.distance_matrix[last_node, j]]) next_minnode = own_pickup_node[next_minIndex] arrival = departure + model.time_matrix[last_node, next_minnode] if arrival <= model.demand_dict[next_minnode].end_time and arrival <= depot.dend_time: route.append(next_minnode) departure = arrival + service_time route.append(own_delivery_node[next_minIndex]) arrival = departure + model.time_matrix[next_minnode, own_delivery_node[next_minIndex]] departure += arrival + service_time last_node = own_delivery_node[next_minIndex] own_pickup_node.remove(next_minnode) own_delivery_node.remove(own_delivery_node[next_minIndex]) else: continue route.append(depot.depot_id) sol.route_list.append(route) vehicle_number = vehicle_number - 1 route = [] print(sol.route_list) return sol.route_list 这段代码的问题是有可能vehicle_number为0了，但是owner_pickup_node的长度还不为0，这种情况怎么解决

这种情况下，程序会陷入无限循环，因为当所有车辆都被分配任务后，仍有剩余的pickup_node，但是vehicle_number为0，所以不会再有车辆来接这些节点。为了解决这个问题，可以在while循环条件中添加一个限制条件，即当所有车辆都被分配任务后，停止while循环。例如可以这样修改while循环的条件：

while vehicle_number > 0 and len(own_pickup_node) > 0 and len(sol.route_list) < model.depot_dict['d1'].depot_capacity:

这样，当所有车辆都被分配任务后，循环就会停止，避免了程序陷入无限循环的情况。

def calCrowdcarCost(crowd_route_list, model): cost_of_distance = 0 crowd_wait_time = 0 model.crowdcarcost = 0 for route in crowd_route_list: timetable = [] for i in range(len(route)): if i == 0: depot_id = route[i] next_node_id = route[i+1] cost_of_distance = model.distance_matrix[depot_id, next_node_id] travel_time = int(model.distance_matrix[depot_id, next_node_id]) departure = max(0, model.demand_dict[next_node_id].start_time - travel_time) if departure == 0: crowd_wait_time = travel_time - model.demand_dict[next_node_id].start_time timetable.append((departure, departure)) elif 1 <= i <= len(route)-2: last_node_id = route[i-1] current_node_id = route[i] current_node = model.demand_dict[current_node_id] travel_time = int(model.distance_matrix[last_node_id, current_node_id]) cost_of_distance += model.distance_matrix[last_node_id, current_node_id] arrival = max(timetable[-1][1]+travel_time, current_node.start_time) if arrival == timetable[-1][1]+travel_time: crowd_wait_time += timetable[-1][1]+travel_time - current_node.start_time else: break model.crowdcarcost += model.fixcost + model.percost * cost_of_distance return model.crowdcarcost, crowd_wait_time 在这一段代码中我想知道每个点的等待时间总和，等待时间等于到达这个的时间减去它的时间窗上限

在这段代码中，你已经计算了每个点的等待时间，它们被累加到了`crowd_wait_time`变量中。如果你想得到每个点的等待时间总和，可以在循环中添加一个变量来记录。以下是修改后的代码：

def calCrowdcarCost(crowd_route_list, model):
    cost_of_distance = 0
    crowd_wait_time = 0
    point_wait_time = {}  # 记录每个点的等待时间
    model.crowdcarcost = 0
    
    for route in crowd_route_list:
        timetable = []
        for i in range(len(route)):
            if i == 0:
                depot_id = route[i]
                next_node_id = route[i+1]
                cost_of_distance = model.distance_matrix[depot_id, next_node_id]
                travel_time = int(model.distance_matrix[depot_id, next_node_id])
                departure = max(0, model.demand_dict[next_node_id].start_time - travel_time)
                if departure == 0:
                    crowd_wait_time = travel_time - model.demand_dict[next_node_id].start_time
                    point_wait_time[next_node_id] = travel_time - model.demand_dict[next_node_id].start_time
                timetable.append((departure, departure))
            elif 1 <= i <= len(route)-2:
                last_node_id = route[i-1]
                current_node_id = route[i]
                current_node = model.demand_dict[current_node_id]
                travel_time = int(model.distance_matrix[last_node_id, current_node_id])
                cost_of_distance += model.distance_matrix[last_node_id, current_node_id]
                arrival = max(timetable[-1][1]+travel_time, current_node.start_time)
                if arrival == timetable[-1][1]+travel_time:
                    wait_time = timetable[-1][1]+travel_time - current_node.start_time
                    crowd_wait_time += wait_time
                    point_wait_time[current_node_id] = wait_time
                else:
                    break
    
    model.crowdcarcost += model.fixcost + model.percost * cost_of_distance
    
    return model.crowdcarcost, crowd_wait_time, point_wait_time

这样，`point_wait_time`字典中的每个键值对代表一个点的等待时间，其中键为点的ID，值为该点的等待时间。