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的搜索算法，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，这是我到目前为止所做的。

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}

对于列表中的所有顶点显示访问顺序为1，当它遍历＆＃34;时，它应该将访问顺序显示为每个顶点的不同数字。＆＃34;我相信这个错误源于bfs（）函数的while循环。尝试修复代码的任何建议，以便我可以实现所需的输出？我也不熟悉Python中的队列，所以感谢任何帮助。

Answer 1

您的代码中存在许多问题 -

首先，您永远不会在您创建的队列中放置任何内容，因此它始终为空，您需要在v循环之前将while放入队列中，是起点。
其次，在for循环中，您正在检查element == 0，这是错误的，您需要检查是否graphv[element] == 0，即该元素是否已被访问过或不。
第三，在for循环中，您需要设置graphv[element] = count，表示您已成功element。
您没有使用 - visited.put()将任何内容放入队列中，您需要将该元素作为参数传递到队列中。
从队列中取回元素时，您需要将其分配回v，否则v永远不会更改，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） -

{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

您可以使用其中一个提供搜索算法的图表库，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

2 个答案: