广度优先搜索算法

Question

就像我之前遇到的一个问题一样，我正在尝试创建一个广度优先搜索算法，该算法采用图形并输出顶点访问顺序。它需要一个邻接矩阵（代表图形）作为输入，这是我目前所拥有的。

import sys
import Queue

# Input has to be adjacency matrix or list
graphAL2 = {0 : [1,2,3],
        1 : [0,3,4],
        2 : [0,4,5],
        3 : [0,1,5],
        4 : [1,2],
        5 : [2,3] }

# NEED TO FIX:
# - Final graphAL2v print is only displaying key values as 1, not iterating
# through graph and visiting each vertex

def main():
    count = 0
    graphAL2v = {}

    for key, value in graphAL2.items():
        graphAL2v[key] = 0

    print(graphAL2v)

    for key in graphAL2v: # each vertex v in V
        if graphAL2v[key] == 0: # is marked with 0
            bfs(key, count, graphAL2, graphAL2v)
    print(graphAL2v)

def bfs(v, count, graphal, graphv):
    count = count + 1
    print('Visiting', v)

    # Mark v with count and initialize queue with v
    graphv[v] = count
    visited = Queue.Queue()

    while not visited.empty(): #queue not empty:
        print('queue is not empty')
        for element in graphal[v]: # each vertex w in V adjacent to front vertex
            if element == 0:
                count = count + 1
                # mark w with count
                graphal[v] = count
                visited.put()
        visited.get()

if __name__ == '__main__':
    sys.exit(main())

我运行遇到的问题是我的输出

{0: 0, 1: 0, 2: 0, 3: 0, 4: 0, 5: 0}
('Visiting', 0)
('Visiting', 1)
('Visiting', 2)
('Visiting', 3)
('Visiting', 4)
('Visiting', 5)
{0: 1, 1: 1, 2: 1, 3: 1, 4: 1, 5: 1}

将列表中所有顶点的访问顺序显示为 1，而当它遍历 "graph." 时应该将每个顶点的访问顺序显示为不同的数字我相信这个错误源于内部bfs() 函数的 while 循环。有关尝试修复代码以便获得所需输出的任何建议吗？我也不太熟悉 Python 中的队列，因此不胜感激。

Answer 1

你的代码中有很多问题 -

首先，你永远不会在你创建的队列中放入任何东西，所以它总是空的，你需要在 [=13= 之前将 v 放入队列中]循环，即起点。
其次，在for循环中，你正在检查element == 0，这是错误的，你需要检查是否graphv[element] == 0，即元素是否是否已经访问过。
第三，在for循环中，你需要设置graphv[element] = count，这表示你活了element.
您没有使用 - visited.put() 将任何内容放入队列中，您需要将要放入队列中的元素作为参数传递。
从Queue取回元素时，需要重新赋值给v，否则v永远不会改变，v表示当前正在迭代的元素。

示例代码-

import sys
import Queue

# Input has to be adjacency matrix or list
graphAL2 = {0 : [1,2,3],
        1 : [0,3,4],
        2 : [0,4,5],
        3 : [0,1,5],
        4 : [1,2],
        5 : [2,3] }

# NEED TO FIX:
# - Final graphAL2v print is only displaying key values as 1, not iterating
# through graph and visiting each vertex

def main():
    count = 0
    graphAL2v = {}

    for key, value in graphAL2.items():
        graphAL2v[key] = 0

    print(graphAL2v)

    for key in graphAL2v: # each vertex v in V
        if graphAL2v[key] == 0: # is marked with 0
            bfs(key, count, graphAL2, graphAL2v)
    print(graphAL2v)

def bfs(v, count, graphal, graphv):
    count = count + 1
    print('Visiting', v)

    # Mark v with count and initialize queue with v
    graphv[v] = count
    visited = Queue.Queue()
    visited.put(v)
    while not visited.empty(): #queue not empty:
        print('queue is not empty')
        for element in graphal[v]: # each vertex w in V adjacent to front vertex
            if graphv[element] == 0:
                count = count + 1
                # mark w with count
                graphv[element] = count
                visited.put(element)
        v = visited.get()
    return count

if __name__ == '__main__':
    sys.exit(main())

演示（经过上述更改后）-

{0: 0, 1: 0, 2: 0, 3: 0, 4: 0, 5: 0}
Visiting 0
queue is not empty
queue is not empty
queue is not empty
queue is not empty
queue is not empty
queue is not empty
{0: 1, 1: 2, 2: 3, 3: 4, 4: 5, 5: 6}

Answer 2

您可以使用提供搜索算法的图形库之一，例如 NetworkX：

from networkx import *
graphAL2 = {0 : [1,2,3],
        1 : [0,3,4],
        2 : [0,4,5],
        3 : [0,1,5],
        4 : [1,2],
        5 : [2,3] }

g = Graph()

# create graph
for node in graphAL2:
    g.add_node(node)
    for target_node in graphAL2[node]:
        g.add_edge(node, target_node)

print bfs_successors(g, 0)

这将获得每个节点的后继者，从中导出搜索顺序应该是小菜一碟。

输出：

{0: [1, 2, 3], 1: [4], 2: [5]}

广度优先搜索算法

Breadth-first search algorithm

python

algorithm

recursion

queue

breadth-first-search