问题29

这是一个关于图的分支问题。让我们看一下这个问题。

问题描述

给定一个有向无环图，其中节点可以分为三个部分：a，b和c。a和b之间没有边，每个节点都至少有一个出边。现在我们想要从a到c，同时满足要求：

• 如果从a开始沿任意路径到达了b，那么必须遵循从b到c的路径，无论哪条路径最终都必须从b到达c。
• 如果从a开始一直到达c的路径不经过b，那么这条路径必须只包含从a到c的边。

解决方案

这是一个典型的拓扑排序问题，我们可以通过如下步骤求解：

1. 首先解决b的位置，假设b的位置为p。遍历所有从a到c的路径，如果路径上没有p，则什么也不必做，否则在p处将路径拆成两段，即从a到p和从p到c。
2. 对于第一段，我们只需要从a开始按拓扑序遍历到p，然后记录每个节点的到达距离（即从a到该节点的路径长度）。这些距离将被用于找到第二段路径时的起点。
3. 对于第二段，我们按照拓扑序从p开始遍历到c，每当经过一个节点时，我们检查是否已经有从该节点开始的路径。如果是这样，我们需要将该节点关联到路径最短的那个起点。

以下是一个实现拓扑排序的Python代码：

def topological_sort(graph):
    in_degree = {u: 0 for u in graph}
    for u in graph:
        for v in graph[u]:
            in_degree[v] += 1
    queue = [u for u in graph if in_degree[u] == 0]
    result = []
    while queue:
        u = queue.pop(0)
        result.append(u)
        for v in graph[u]:
            in_degree[v] -= 1
            if in_degree[v] == 0:
                queue.append(v)
    return result

def shortest_paths(graph, start):
    distances = {v: float('inf') for v in graph}
    distances[start] = 0
    queue = [start]
    while queue:
        u = queue.pop(0)
        for v in graph[u]:
            if distances[v] > distances[u] + 1:
                distances[v] = distances[u] + 1
                queue.append(v)
    return distances

def find_paths(graph, start, end):
    stack = [(start, [])]
    paths = []
    while stack:
        (vertex, path) = stack.pop()
        for next_vertex in (set(graph[vertex]) - set(path)):
            if next_vertex == end:
                paths.append(tuple(path + [vertex, next_vertex]))
            else:
                stack.append((next_vertex, path + [vertex]))
    return paths

def split_paths(paths, p):
    paths_a_to_p = set()
    paths_p_to_c = set()
    for path in paths:
        if p not in path:
            paths_a_to_p.add(path)
        else:
            index = path.index(p)
            paths_a_to_p.add(path[:index+1])
            paths_p_to_c.add(path[index:])
    return paths_a_to_p, paths_p_to_c

def associate_paths(paths, distances):
    associated = {}
    for path in paths:
        if path[0] not in distances:
            continue
        path_distance = sum(distances[v] for v in path[:-1])
        if path[-1] not in associated or associated[path[-1]][1] > path_distance:
            associated[path[-1]] = (path, path_distance)
    return associated

def find_shortest_association(paths, distances):
    associations = associate_paths(paths, distances)
    if not associations:
        return None
    min_distance = min(association[1] for association in associations.values())
    for node in sorted(associations.keys()):
        if associations[node][1] == min_distance:
            return associations[node][0]

def solve(graph, a, b, c):
    sorted_nodes = topological_sort(graph)
    p_index = sorted_nodes.index(b)
    paths = find_paths(graph, a, c)
    paths_a_to_p, paths_p_to_c = split_paths(paths, b)
    distances = shortest_paths(graph, a)
    for path in paths_a_to_p:
        for node in path:
            if sorted_nodes.index(node) >= p_index:
                break
            distances[node] = max(distances[node], distances[b])
    return find_shortest_association(paths_p_to_c, distances)

测试样例

我们使用以下样例对上述实现进行测试：

graph = {
    'a': ['d', 'e'],
    'b': ['e', 'f'],
    'c': ['f', 'g', 'h'],
    'd': ['i'],
    'e': ['i'],
    'f': ['j'],
    'g': ['j'],
    'h': ['j'],
    'i': ['k'],
    'j': ['k'],
    'k': ['c']
}

assert solve(graph, 'a', 'b', 'c') == ('a', 'e', 'i', 'k', 'c')