Question

我有一个二分加权图（更确切地说是一个分配问题），我想找到最便宜的路径/选择。知道如何实现DFS或Greedy BFS来实现这条路径吗？

我将图表表示为一个adiacency列表（或者我认为最好），就像这样和DFS算法

adjlist = {"A": ["W", "X"],
           "B": ["W", "X", "Y", "Z"],
           "C": ["Y", "Z"],
           "D": ["W", "X", "Y", "Z"],
           "W": ["A", "B", "D"],
           "X": ["A", "B", "D"],
           "Y": ["B", "C", "D"],
           "Z": ["B", "C", "D"] }

def dfser(graph, root):
    visited =[]
    stack = [root, ]

    while stack:
        node = stack.pop()
        if node not in visited:
            visited.append(node)
            stack.extend([x for x in graph[node] if x not in visited])
    return visited

我想要的可能吗？结果必须是：AWBXCYDZ或最便宜的东西。

或者我可以从root获得所有可能的遍历吗？这个DFS只给我一个，但我想要所有可能的转换

Answer 1

BFS或DFS不适用于最短路径搜索，尤其是在加权图表中。

使用dijkstra或A*算法可以查找最便宜的路径。 Networkx为该任务提供full API。

networkx模块实现bipartite graphs及相关算法：

Dicti = {"A": {"W": 2, "X": 2},
     "B": {"W": 4, "X": 3, "Y": 2, "Z": 5},
     "C": {"Y": 2, "Z": 2},
     "D": {"W": 5, "X": 5, "Y": 4, "Z": 3},
     "W": {"A": 2, "B": 4, "D": 5},
     "X": {"A": 2, "B": 3, "D": 5},
     "Y": {"B": 2, "C": 2, "D": 4},
     "Z": {"B": 5, "C": 2, "D": 3}
}

graph = nx.Graph()
for node, succs in Dicti.items():
    for succ, weight in succs:
        graph.add_edge(node, succ, weight=weight)

print(nx.dijkstra_path(graph, 'A', 'Z'))  # here is Dijkstra

在二分加权图DFS / Greedy BFS中最便宜的路径

1 个答案: