利用Breadth-First Search (BFS)算法寻找图中的最短路径和所有路径

       今天在stackoverflow网站搜索问题时，发现了一个用BFS算法搜索图中最短路径比较简洁且容易理解的代码。暂且放在博客记录下来，方便今后用到。

      如上图，我们要使用BFS算法搜索1—11的最短路径，代码如下：

# graph is in adjacent list representationgraph = {        '1': ['2', '3', '4'],        '2': ['5', '6'],        '5': ['9', '10'],        '4': ['7', '8'],        '7': ['11', '12']        }def bfs(graph, start, end):    # maintain a queue of paths    queue = []    # push the first path into the queue    queue.append([start])    while queue:        # get the first path from the queue        path = queue.pop(0)        # get the last node from the path        node = path[-1]        # path found        if node == end:            return path        # enumerate all adjacent nodes, construct a new path and push it into the queue        for adjacent in graph.get(node, []):            new_path = list(path)            new_path.append(adjacent)            queue.append(new_path)print bfs(graph, '1', '11')

输出结果：[1，4，7，11]

      如果我们修改几个地方，增加了节点3的临节点，网络拓扑已改变。则可以得到源和目的节点的所有路径，代码如下：

# graph is in adjacent list representationgraph = {        1: [2, 3, 4],        2: [5, 6],        3: [4,11],        5: [9, 10],        4: [7, 8],        7: [11, 12]        }def bfs(graph, start, end):    # maintain a queue of paths    queue = []    allpath = []    # push the first path into the queue    queue.append([start])    while queue:        # get the first path from the queue        path = queue.pop(0)        # get the last node from the path        node = path[-1]        # path found        if node == end:            allpath.append(path)        # enumerate all adjacent nodes, construct a new path and push it into the queue        for adjacent in graph.get(node, []):            new_path = list(path)            new_path.append(adjacent)            queue.append(new_path)    return allpathprint bfs(graph, 1, 11)

输出结果：[[1, 3, 11], [1, 4, 7, 11], [1, 3, 4, 7, 11]]