python - 使用字典定义图形并查找路径

Question

我已经定义了如下图：

adj_matrix = {'1': set('2'),
              '2': set('3'),
              '3': set(['4', '5']),
              '4': set(''),
              '5': set('6'),
              '6': set('7'),
              '7': set('8'),
              '8': set(['9', '14']),
              '9': set(['10', '11']),
              '10': set(''),
              '11': set(['12', '13']),
              '12': set(''),
              '13': set(''),
              '14': set('15'),
              '15': set('16'),
              '16': set('17'),
              '17': set(['18', '19']),
              '18': set(''),
              '19': set('')}

我正在使用以下内容来查找不同节点之间的路径：

def find_path(graph, start, end, path=[]):
        path = path + [start]
        if start == end:
            return path
        if start not in graph:
            return None
        for node in graph[start]:
            if node not in path:
                newpath = find_path(graph, node, end, path)
                if newpath: return newpath
        return None

问题是该功能仅适用于节点1到14 。 就像我称之为p = find_path(adj_matrix, '3', '14')它就像一个魅力，给我['3', '5', '6', '7', '8', '14']作为答案，但是当我尝试p = find_path(adj_matrix, '3', '15') ，它返回None 。 任何想法发生了什么以及如何解决这个问题？

Answer 1

对于错误的情况，当你达到'14' ，可用的路线显示为'1'和'5' ，而不是'15' 。 您应该为每个set设置一个列表，以便您可以迭代实际值。

adj_matrix = {'1': set(['2']),
              '2': set(['3']),
              '3': set(['4', '5']),
              '4': set(''),
              '5': set(['6']),
              '6': set(['7']),
              '7': set(['8']),
              '8': set(['9', '14']),
              '9': set(['10', '11']),
              '10': set(''),
              '11': set(['12', '13']),
              '12': set(''),
              '13': set(''),
              '14': set(['15']),
              '15': set(['16']),
              '16': set(['17']),
              '17': set(['18', '19']),
              '18': set(''),
              '19': set('')}

使用上面的adj_matrix，它给出了正确的结果。

这是我的调试器的功能。 您可以看到代码如何从终端流出：

def find_path(graph, start, end, path=[]):
        path = path + [start]
        print("st:%s, end:%s, path:%s" % (start,end,path))
        if start == end:
            return path
        if start not in graph:
            return None
        print("available routes: %s" % graph[start])
        for node in graph[start]:
            if node not in path:
                print("next node: %s" % node)
                newpath = find_path(graph, node, end, path)
                if newpath: return newpath
        return None

使用你错误的adj_matrix，它会给出如下控制台输出:(关注'14'时要仔细观察）

C:\Users\rapac\Desktop\stackoverflow>python adjgraph.py
st:3, end:15, path:['3']
available routes: {'5', '4'}
next node: 5
st:5, end:15, path:['3', '5']
available routes: {'6'}
next node: 6
st:6, end:15, path:['3', '5', '6']
available routes: {'7'}
next node: 7
st:7, end:15, path:['3', '5', '6', '7']
available routes: {'8'}
next node: 8
st:8, end:15, path:['3', '5', '6', '7', '8']
available routes: {'9', '14'}
next node: 9
st:9, end:15, path:['3', '5', '6', '7', '8', '9']
available routes: {'11', '10'}
next node: 11
st:11, end:15, path:['3', '5', '6', '7', '8', '9', '11']
available routes: {'13', '12'}
next node: 13
st:13, end:15, path:['3', '5', '6', '7', '8', '9', '11', '13']
available routes: set()
next node: 12
st:12, end:15, path:['3', '5', '6', '7', '8', '9', '11', '12']
available routes: set()
next node: 10
st:10, end:15, path:['3', '5', '6', '7', '8', '9', '10']
available routes: set()
next node: 14
st:14, end:15, path:['3', '5', '6', '7', '8', '14']
available routes: {'5', '1'} <--- Here it is! We expect '15' here.
next node: 1
st:1, end:15, path:['3', '5', '6', '7', '8', '14', '1']
available routes: {'2'}
next node: 2
st:2, end:15, path:['3', '5', '6', '7', '8', '14', '1', '2']
available routes: {'3'}
next node: 4
st:4, end:15, path:['3', '4']
available routes: set()
None

最后，这是使用正确版本的adj_matrix时的输出：

C:\Users\rapac\Desktop\stackoverflow>python adjgraph.py
st:3, end:15, path:['3']
available routes: {'4', '5'}
next node: 4
st:4, end:15, path:['3', '4']
available routes: set()
next node: 5
st:5, end:15, path:['3', '5']
available routes: {'6'}
next node: 6
st:6, end:15, path:['3', '5', '6']
available routes: {'7'}
next node: 7
st:7, end:15, path:['3', '5', '6', '7']
available routes: {'8'}
next node: 8
st:8, end:15, path:['3', '5', '6', '7', '8']
available routes: {'9', '14'}
next node: 9
st:9, end:15, path:['3', '5', '6', '7', '8', '9']
available routes: {'11', '10'}
next node: 11
st:11, end:15, path:['3', '5', '6', '7', '8', '9', '11']
available routes: {'12', '13'}
next node: 12
st:12, end:15, path:['3', '5', '6', '7', '8', '9', '11', '12']
available routes: set()
next node: 13
st:13, end:15, path:['3', '5', '6', '7', '8', '9', '11', '13']
available routes: set()
next node: 10
st:10, end:15, path:['3', '5', '6', '7', '8', '9', '10']
available routes: set()
next node: 14
st:14, end:15, path:['3', '5', '6', '7', '8', '14']
available routes: {'15'} <-- Now we're good.
next node: 15
st:15, end:15, path:['3', '5', '6', '7', '8', '14', '15']
['3', '5', '6', '7', '8', '14', '15']

现在代码的行为与您期望的一样。 希望能帮助到你。

Answer 2

如果您将输入更改为

adj_matrix = {'1': set('2'),
          '2': set('3'),
          '3': set(['4', '5']),
          '4': set(''),
          '5': set('6'),
          '6': set('7'),
          '7': set('8'),
          '8': set(['9', '14']),
          '9': set(['10', '11']),
          '10': set(''),
          '11': set(['12', '13']),
          '12': set(''),
          '13': set(''),
          '14': set(['15']),
          '15': set(['16']),
          '16': set(['17']),
          '17': set(['18', '19']),
          '18': set(''),
          '19': set('')}

（我刚刚添加了'15'，'16'和'17'列表。）然后它工作。 你的问题与这样一个事实有关：当它们在集合中时，两位数字被解释为for node in graph[start]:行中的一个字符串for node in graph[start]:然后你循环遍历['1'，'5']

另请注意，您可以简单地使用整数而不是每个数字周围的引号，这也可以解决这个问题。