Question

我想找到从goal到root向后工作的最短路径

我对root的输入是{'4345092': ['6570646', '40586', '484']} 我对goal的输入是{'886619': ['GOAL']}

我对path_holder的输入是输入，但它会转换为dct并用于此功能。我对while循环感到困惑，因为它为我创建了向后的路径。现在我无法打印q，因为那部分代码没有运行。 dct基本上是包含循环的有向图表示。我似乎无法弄清楚如何从GOAL开始并最终到达root节点。我想知道是否有人可以帮我解决这个问题！

DCT：

dct = 
{ '612803266': ['12408765', '46589', '5880', '31848'], 
  '8140983': ['7922972', '56008'],
  '7496838': ['612803266'], 
  '1558536111': ['7496838'], 
  '31848': ['DEADEND'], 
  '1910530': ['8140983'], 
  '242010': ['58644', '886619'], 
  '727315568': ['DEADEND'], 
  '12408765': ['DEADEND'], 
  '56008': ['DEADEND'], 
  '58644': ['DEADEND'], 
  '886619': ['GOAL'], 
  '40586': ['931', '727315568', '242010', '1910530'], 
  '5880': ['1558536111'], 
  '46589': ['DEADEND'], 
  '6570646': ['2549003','43045', '13830'],
  '931': ['299159122'],
  '484': ['1311310', '612803266'],
  '1311310': ['DEADEND'],
  '7922972': ['DEADEND']
  }

我的职能：

def backtrace(path_holder, root, goal):
    dct = {}
    for d in path_holder:
            dct.update(d)
    rootnode = root.keys()[0]
    goal = goal.keys()[0]
    path = []
    path.append(goal)
    q = 0
    while goal != rootnode: 
        # find key that contains goal in list
        for i in dct:   #iterate keys
            if i in dct:    # prevent repeat of path
                continue
            for j in dct[i]:    #iterate though children
                if j == goal:
                    path.append(i)
                    goal = i    # look for new goal
                    q += 1
                    print q
                    #print goal

        # append key that has goal in the list
        # set goal to be the key that was appended
        # repeat

    return path

Answer 1

找到路径然后反转它们。

更新：在最终条件中将“[]”添加到“DEADEND”和“GOAL”。

import copy as cp

DCT = {...}  # You already know what goes here. 

FOUND_PATHS = []            # In case of more than one path to GOAL.
FOUND_REVERSE_PATHS = []

COUNTER = len(DCT)

def back_track(root, target_path = [], counter=COUNTER):
    """
    @param root: DCT key.
    @type root: str.

    @param target_path: Reference to the path we are constructing.
    @type target_path: list.

    """
    global FOUND_PATHS

    # Avoiding cycles.
    if counter == 0:
        return

    # Some nodes aren't full generated.
    try:
        DCT[root]
    except KeyError:
        return

    # End condition.
    if DCT[root] == ['DEADEND']:
        return
    # Path found.
    if DCT[root] == ['GOAL']:
        FOUND_PATHS.append(target_path)             # The normal path.
        reverse_path = cp.copy(target_path)
        reverse_path.reverse()
        FOUND_REVERSE_PATHS.append(reverse_path)     # The path you want.
        return

    for node in DCT[root]:
        # Makes copy of target parh and add the node.
        path_copy = cp.copy(target_path)
        path_copy.append(node)
        # Call back_track with current node and the copy 
        # of target_path.
        back_track(node, path_copy, counter=(counter - 1))

if __name__ == '__main__':
    back_track('4345092')
    print(FOUND_PATHS)
    print(FOUND_REVERSE_PATHS)

在循环python的有向图中从目标到根的最短路径

1 个答案: