Question

我有这个python代码来计算Dijkstra算法。如何在终端（ubuntu）中打印最短路径？我尝试了一些打印功能，但他们提出了不同的例外情况...... 提前谢谢！

class Graph(object):
"""
A simple undirected, weighted graph
"""
   def __init__(self):
       self.nodes = set()
       self.edges = {}
       self.distances = {}

   def add_node(self, value):
       self.nodes.add(value)

   def add_edge(self, from_node, to_node, distance):
       self._add_edge(from_node, to_node, distance)
       self._add_edge(to_node, from_node, distance)

   def _add_edge(self, from_node, to_node, distance):
       self.edges.setdefault(from_node, [])
       self.edges[from_node].append(to_node)
       self.distances[(from_node, to_node)] = distance


def dijkstra(graph, initial_node):
    visited = {initial_node: 0}
    current_node = initial_node
    path = {}

    nodes = set(graph.nodes)

    while nodes:
        min_node = None
        for node in nodes:
            if node in visited:
                if min_node is None:
                    min_node = node
                elif visited[node] < visited[min_node]:
                    min_node = node

        if min_node is None:
            break

        nodes.remove(min_node)
        cur_wt = visited[min_node]

        for edge in graph.edges[min_node]:
            wt = cur_wt + graph.distances[(min_node, edge)]
            if edge not in visited or wt < visited[edge]:
                visited[edge] = wt
                path[edge] = min_node

    return visited, path

def route(graph, x, y):
    distances, paths = dijkstra(graph, x)
    route = [y]

    while y != x:
        route.append(paths[y])
        y = paths[y]

    route.reverse()
    return route



if __name__ == '__main__':
    g = Graph()
    g.nodes = set(range(1, 7))
    g.add_edge(1, 2, 7)
    g.add_edge(1, 3, 9)
    g.add_edge(1, 6, 14)
    g.add_edge(2, 3, 10)
    g.add_edge(2, 4, 15)
    g.add_edge(3, 4, 11)
    g.add_edge(3, 6, 2)
    g.add_edge(4, 5, 6)
    g.add_edge(5, 6, 9)
    assert route(g, 1, 5) == [1, 3, 6, 5]
    assert route(g, 5, 1) == [5, 6, 3, 1]
    assert route(g, 2, 5) == [2, 3, 6, 5]
    assert route(g, 1, 4) == [1, 3, 4]

Answer 1

可视化生成的路径：

def print_route(graph, x, y):
  r = route(graph, x, y)
  prmpt = ['({})'.format(x)]
  for y in r[1:]:
    d = graph.distances.get((x, y))
    prmpt.append(' --{}-> ({})'.format(d,y))
    x = y
  print(''.join(prmpt))

输出节点1和5之间的最短路径：

(1) --9-> (3) --2-> (6) --9-> (5)

打印终端中的最短路径

1 个答案: