具有给定数量要访问的位置的旅行商问题

Question

有一个很好的例子，这里的如何找到一个解决旅行商问题：

"""Simple travelling salesman problem between cities."""

from __future__ import print_function
from ortools.constraint_solver import routing_enums_pb2
from ortools.constraint_solver import pywrapcp



def create_data_model():
    """Stores the data for the problem."""
    data = {}
    data['distance_matrix'] = [
        [0, 2451, 713, 1018, 1631, 1374, 2408, 213, 2571, 875, 1420, 2145, 1972],
        [2451, 0, 1745, 1524, 831, 1240, 959, 2596, 403, 1589, 1374, 357, 579],
        [713, 1745, 0, 355, 920, 803, 1737, 851, 1858, 262, 940, 1453, 1260],
        [1018, 1524, 355, 0, 700, 862, 1395, 1123, 1584, 466, 1056, 1280, 987],
        [1631, 831, 920, 700, 0, 663, 1021, 1769, 949, 796, 879, 586, 371],
        [1374, 1240, 803, 862, 663, 0, 1681, 1551, 1765, 547, 225, 887, 999],
        [2408, 959, 1737, 1395, 1021, 1681, 0, 2493, 678, 1724, 1891, 1114, 701],
        [213, 2596, 851, 1123, 1769, 1551, 2493, 0, 2699, 1038, 1605, 2300, 2099],
        [2571, 403, 1858, 1584, 949, 1765, 678, 2699, 0, 1744, 1645, 653, 600],
        [875, 1589, 262, 466, 796, 547, 1724, 1038, 1744, 0, 679, 1272, 1162],
        [1420, 1374, 940, 1056, 879, 225, 1891, 1605, 1645, 679, 0, 1017, 1200],
        [2145, 357, 1453, 1280, 586, 887, 1114, 2300, 653, 1272, 1017, 0, 504],
        [1972, 579, 1260, 987, 371, 999, 701, 2099, 600, 1162, 1200, 504, 0],
    ]  # yapf: disable
    data['num_vehicles'] = 1
    data['depot'] = 0
    return data


def print_solution(manager, routing, assignment):
    """Prints assignment on console."""
    print('Objective: {} miles'.format(assignment.ObjectiveValue()))
    index = routing.Start(0)
    plan_output = 'Route for vehicle 0:\n'
    route_distance = 0
    while not routing.IsEnd(index):
        plan_output += ' {} ->'.format(manager.IndexToNode(index))
        previous_index = index
        index = assignment.Value(routing.NextVar(index))
        route_distance += routing.GetArcCostForVehicle(previous_index, index, 0)
    plan_output += ' {}\n'.format(manager.IndexToNode(index))
    print(plan_output)
    plan_output += 'Route distance: {}miles\n'.format(route_distance)


def main():
    """Entry point of the program."""
    # Instantiate the data problem.
    data = create_data_model()

    # Create the routing index manager.
    manager = pywrapcp.RoutingIndexManager(len(data['distance_matrix']),
                                           data['num_vehicles'], data['depot'])

    # Create Routing Model.
    routing = pywrapcp.RoutingModel(manager)


    def distance_callback(from_index, to_index):
        """Returns the distance between the two nodes."""
        # Convert from routing variable Index to distance matrix NodeIndex.
        from_node = manager.IndexToNode(from_index)
        to_node = manager.IndexToNode(to_index)
        return data['distance_matrix'][from_node][to_node]

    transit_callback_index = routing.RegisterTransitCallback(distance_callback)

    # Define cost of each arc.
    routing.SetArcCostEvaluatorOfAllVehicles(transit_callback_index)

    # Setting first solution heuristic.
    search_parameters = pywrapcp.DefaultRoutingSearchParameters()
    search_parameters.first_solution_strategy = (
        routing_enums_pb2.FirstSolutionStrategy.PATH_CHEAPEST_ARC)

    # Solve the problem.
    assignment = routing.SolveWithParameters(search_parameters)

    # Print solution on console.
    if assignment:
        print_solution(manager, routing, assignment)


if __name__ == '__main__':
    main()

它返回

Objective: 7293 miles
Route for vehicle 0:
 0 -> 7 -> 2 -> 3 -> 4 -> 12 -> 6 -> 8 -> 1 -> 11 -> 10 -> 5 -> 9 -> 0

我想修改代码，以便我仍然从节点0开始，但不必访问所有节点，我只需要访问其中的 5 个。 如何相应地更改代码？

我无法改变

    manager = pywrapcp.RoutingIndexManager(len(data['distance_matrix']),
                                           data['num_vehicles'], data['depot'])

到

    manager = pywrapcp.RoutingIndexManager(5,
                                           data['num_vehicles'], data['depot'])

因为我仍然想考虑所有节点 - 我只想访问其中的 5 个。

Answer 1

您可以使用分离约束使所有停靠点成为可选，然后引入停靠点计数器维度并将其容量限制为 5。

析取约束：

n_stops = len(data['distance_matrix'])
n2x = index_manager.NodeToIndex
penalty = 100  # don't know if this is a good choice

for node in range(1, n_stops):  # skip depot
    routing_model.AddDisjunction([n2x(node)], penalty)

停止计数器尺寸：

fix_start_cumul_to_zero = True
slack_max = 0
name = 'stop_counter_dimension'
capacity = 5

def evaluator(i, j):
    return 1

callback_index = routing_model.RegisterTransitCallback(evaluator)
routing_model.AddDimension(callback_index, slack_max, capacity, fix_start_cumul_to_zero, name)