组合成对

Question

我的工作有向网络问题，并试图计算两个点之间的所有有效路径。我需要在路径的方式来看待的长度可达30“跳闸”（由[出发地，目的地]对表示）。然后完整的路线由一系列这些对的：

route = [[start, city2], [city2, city3], [city3, city4], [city4, city5], [city5, city6], [city6, city7], [city7, city8], [city8, stop]]

到目前为止我的最佳解决方案是如下所示：

    def numRoutes(graph, start, stop, minStops, maxStops):
    routes = []

    route = [[start, stop]]
    if distance(graph, route) != "NO SUCH ROUTE" and len(route) >= minStops and len(route) <= maxStops: 
            routes.append(route)

    if maxStops >= 2:
        for city2 in routesFromCity(graph, start):
            route = [[start, city2],[city2, stop]]
            if distance(graph, route) != "NO SUCH ROUTE" and len(route) >= minStops and len(route) <= maxStops: 
                routes.append(route)
    if maxStops >= 3:       
        for city2 in routesFromCity(graph, start):
            for city3 in routesFromCity(graph, city2):
                route = [[start, city2], [city2, city3], [city3, stop]]
                if distance(graph, route) != "NO SUCH ROUTE" and len(route) >= minStops and len(route) <= maxStops:
                    routes.append(route)

    if maxStops >= 4:
        for city2 in routesFromCity(graph, start):
            for city3 in routesFromCity(graph, city2):
                for city4 in routesFromCity(graph, city3):
                    route = [[start, city2], [city2, city3], [city3, city4], [city4, stop]]
                    if distance(graph, route) != "NO SUCH ROUTE" and len(route) >= minStops and len(route) <= maxStops:
                        routes.append(route)
    if maxStops >= 5:
        for city2 in routesFromCity(graph, start):
            for city3 in routesFromCity(graph, city2):
                for city4 in routesFromCity(graph, city3):
                    for city5 in routesFromCity(graph, city4):
                        route = [[start, city2], [city2, city3], [city3, city4], [city4, city5], [city5, stop]]
                        if distance(graph, route) != "NO SUCH ROUTE" and len(route) >= minStops and len(route) <= maxStops:
                            routes.append(route)
return routes

其中numRoutes被馈送我的网络图，其中数字代表的距离：

[[0, 5, 0, 5, 7], [0, 0, 4, 0, 0], [0, 0, 0, 8, 2], [0, 0, 8, 0, 6], [0, 3, 0, 0, 0]]

开始的城市，结束城市和用于路由的长度的参数。

距离检查是否一个路线是可行和routesFromCity返回相连的节点到每个城市馈电。

我有一种感觉，有一个更有效的方式来产生所有的路线，尤其是当我向更多的步骤移动，但我似乎无法得到别的工作了。

Answer 1

您可以使用递归函数。您的maxStops可以是一个参数，并在每次打电话给你减1此时当minStops为0，则得到一个结果，当maxStops是0你停止递归。

下面是一个代码示例：

def routesFromCity(x):
    for i in range(2, 10):
        yield x * i

def findRoutes(start, stop, minStops, maxStops):
    if start == stop:
        if minStops <= 0:
            yield []
    else:
        if maxStops > 0:
            for nextCity in routesFromCity(start):
                for route in findRoutes(nextCity, stop, minStops - 1, maxStops - 1):
                    yield [(start, nextCity)] + route

for route in findRoutes(1, 12, 2, 5):
    print route

输出：

[(1, 2), (2, 4), (4, 12)]
[(1, 2), (2, 6), (6, 12)]
[(1, 2), (2, 12)]
[(1, 3), (3, 6), (6, 12)]
[(1, 3), (3, 12)]
[(1, 4), (4, 12)]
[(1, 6), (6, 12)]