python数据结构与算法——图的广度优先和深度优先的算法

根据维基百科的伪代码实现：

广度优先BFS：

使用队列，集合

标记初始结点已被发现，放入队列

每次循环从队列弹出一个结点

将该节点的所有相连结点放入队列，并标记已被发现

通过队列，将迷宫路口所有的门打开，从一个门进去继续打开里面的门，然后返回前一个门处

 1 """
 2  procedure BFS(G,v) is
 3      let Q be a queue
 4      Q.enqueue(v)
 5      label v as discovered
 6      while Q is not empty
 7         v ← Q.dequeue()
 8         procedure(v)
 9         for all edges from v to w in G.adjacentEdges(v) do
10             if w is not labeled as discovered
11                 Q.enqueue(w)
12                 label w as discovered
13 """
14 def procedure(v):
15     pass
16
17 def BFS(G,v0):
18     """ 广度优先搜索 """
19     q, s = [], set()
20     q.extend(v0)
21     s.add(v0)
22     while q:    # 当队列q非空
23         v = q.pop(0)
24         procedure(v)
25         for w in G[v]:     # 对图G中顶点v的所有邻近点w
26             if w not in s: # 如果顶点 w 没被发现
27                 q.extend(w)
28                 s.add(w)    # 记录w已被发现

深度优先DFS

使用 栈，集合

初始结点入栈

每轮循环从栈中弹出一个结点，并标记已被发现

对每个弹出的结点，将其连接的所有结点放到队列中

通过栈的结构，一步步深入挖掘

 1 """"
 2 Pseudocode[edit]
 3 Input: A graph G and a vertex v of G
 4
 5 Output: All vertices reachable from v labeled as discovered
 6
 7 A recursive implementation of DFS:[5]
 8
 9 1  procedure DFS(G,v):
10 2      label v as discovered
11 3      for all edges from v to w in G.adjacentEdges(v) do
12 4          if vertex w is not labeled as discovered then
13 5              recursively call DFS(G,w)
14 A non-recursive implementation of DFS:[6]
15
16 1  procedure DFS-iterative(G,v):
17 2      let S be a stack
18 3      S.push(v)
19 4      while S is not empty
20 5            v = S.pop()
21 6            if v is not labeled as discovered:
22 7                label v as discovered
23 8                for all edges from v to w in G.adjacentEdges(v) do
24 9                    S.push(w)
25 """
26
27 def DFS(G,v0):
28     S = []
29     S.append(v0)
30     label = set()
31     while S:
32         v = S.pop()
33         if v not in label:
34             label.add(v)
35             procedure(v)
36             for w in G[v]:
37                 S.append(w)