2017-03-09 106 views
1

我试图在Python中运行Dijkstra算法。但是,当我在Eclipse下执行此代码时,没有显示任何内容。我知道pydev能够理解我所有的进口产品,并且没有任何错误信息。任何人都可以指出我正确的方向吗?为什么当我运行我的程序时什么都没有发生

class Graph(object): 
    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 shortest_path(graph, initial_node, goal_node): 
    distances, paths = dijkstra(graph, initial_node) 
    route = [goal_node] 

    while goal_node != initial_node: 
     route.append(paths[goal_node]) 
     goal_node = paths[goal_node] 

    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 shortest_path(g, 1, 5) == [1, 3, 6, 5] 
    assert shortest_path(g, 5, 1) == [5, 6, 3, 1] 
    assert shortest_path(g, 2, 5) == [2, 3, 6, 5] 
    assert shortest_path(g, 1, 4) == [1, 3, 4] 
+0

你预计会发生什么?你没有任何输出。 – Raz

回答

0

假设g是一个图形对象,您可以通过以下尝试打印节点,边和距离:

print g.nodes 
print g.edges 
print g.distances 

它将帮助您了解图形结构,并尝试使用打印,而不是断言的。 在你的情况,这三条线将打印如下:

set([1, 2, 3, 4, 5, 6]) 
{1: [2, 3, 6], 2: [1, 3, 4], 3: [1, 2, 4, 6], 4: [2, 3, 5], 5: [4, 6], 6:[1, 3, 5]} 
{(1, 2): 7, (3, 2): 10, (1, 3): 9, (3, 6): 2, (4, 5): 6, (6, 1): 14, (3, 1): 9, (5, 4): 6, (2, 1): 7, (6, 3): 2, (5, 6): 9, (1, 6): 14, (4, 3): 11, (4, 2): 15, (2, 3): 10, (3, 4): 11, (2, 4): 15, (6, 5): 9} 

希望它能帮助!

0

您未打印任何内容。如果声明不失败,它不会显示任何内容。

而不是assert使用print知道shortest_path是否返回这些数组。 (你实际上可以知道这是因为assert不会引发任何错误)。

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