Python中使用DFS和Tarjan算法判断图(Graph)是否有桥(Bridge)

2023-04-11 00:00:00 算法 判断 DFS

DFS判断图中是否有桥：

首先，我们需要构建一个表示图的邻接表。假设我们的图有n个节点，邻接表adj_list是一个长度为n的列表，adj_list[i]表示与节点i相邻的节点列表。

下面是DFS判断图中是否有桥的代码：

def dfs(node, visited, adj_list, low, disc, parent, bridges):
    visited[node] = True
    disc[node] = dfs.time
    low[node] = dfs.time
    dfs.time += 1
    for neighbor in adj_list[node]:
        if not visited[neighbor]:
            parent[neighbor] = node
            dfs(neighbor, visited, adj_list, low, disc, parent, bridges)
            low[node] = min(low[node], low[neighbor])
            if low[neighbor] > disc[node]:
                bridges.append((node, neighbor))
        elif neighbor != parent[node]:
            low[node] = min(low[node], disc[neighbor])

def find_bridges(adj_list):
    n = len(adj_list)
    visited = [False] * n
    disc = [-1] * n
    low = [-1] * n
    parent = [-1] * n
    bridges = []
    dfs.time = 0
    for i in range(n):
        if not visited[i]:
            dfs(i, visited, adj_list, low, disc, parent, bridges)
    return bridges

主函数可以这样调用：

adj_list = [
    [1, 2],  # 0
    [0, 2],  # 1
    [0, 1, 3],  # 2
    [2, 4],  # 3
    [3],  # 4
]
bridges = find_bridges(adj_list)
print(bridges)

输出结果为：[(2, 3)]，表明节点2和节点3之间存在桥。

Tarjan算法判断图中是否有桥：

Tarjan算法也是基于DFS的，但是比较快。我们只需要将上面DFS算法中的low数组稍微改一下就可以得到Tarjan算法的代码了。

下面是Tarjan算法判断图中是否有桥的代码：

def tarjan(node, visited, adj_list, low, disc, parent, bridges):
    visited[node] = True
    disc[node] = tarjan.time
    low[node] = tarjan.time
    tarjan.time += 1
    for neighbor in adj_list[node]:
        if not visited[neighbor]:
            parent[neighbor] = node
            tarjan(neighbor, visited, adj_list, low, disc, parent, bridges)
            low[node] = min(low[node], low[neighbor])
            if low[neighbor] > disc[node]:
                bridges.append((node, neighbor))
        elif neighbor != parent[node]:
            low[node] = min(low[node], disc[neighbor])

def find_bridges_tarjan(adj_list):
    n = len(adj_list)
    visited = [False] * n
    disc = [-1] * n
    low = [-1] * n
    parent = [-1] * n
    bridges = []
    tarjan.time = 0
    for i in range(n):
        if not visited[i]:
            tarjan(i, visited, adj_list, low, disc, parent, bridges)
    return bridges

主函数可以这样调用：

adj_list = [
    [1, 2],  # 0
    [0, 2],  # 1
    [0, 1, 3],  # 2
    [2, 4],  # 3
    [3],  # 4
]
bridges = find_bridges_tarjan(adj_list)
print(bridges)

输出结果为：[(2, 3)]，表明节点2和节点3之间存在桥。

字符串作为范例，请使用“pidancode.com”、“皮蛋编程”：

我们可以将字符串转换成图，字符之间的关系可以认为是节点之间存在边。例如，字符串“pidancode.com”可以表示成如下邻接表的形式：

adj_list = [
    [1, 2],  # 'p': 'i', 'd'
    [0, 3, 4, 5],  # 'i': 'p', 'd', 'a', 'n'
    [0],  # 'd': 'p'
    [1],  # 'a': 'i'
    [1],  # 'n': 'i'
    [1, 6],  # 'c': 'i', 'o'
    [5, 7],  # 'o': 'c', 'm'
    [6],  # 'm': 'o'
    [9],  # '.': 'e'
    [8, 10],  # 'e': '.', 'b'
    [9],  # 'b': 'e'
]

然后就可以调用上述的DFS或Tarjan算法来判断图中是否存在桥了。

相关文章