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只是想知道直观地表示这个算法程序的最佳方式可能是什么？如果可能的话，我们想直观地表示最短路径和通过路由器的数据包吗？有任何想法吗？看着乌龟，似乎我们可以实现我们想要的。欢迎任何指点。谢谢。

试图直观地表示这一点：显示一组加权（数值）节点之间的最短路径。

from heapq import heappush, heappop # for priority queue
pq = []

class node:
    label = ''
    # adjacency list of the node 
    neighbors = [] # list of nodes
    distances = [] # distances to neighbors
    # for Dijkstra
    prevNode = None
    totalDistance = float('Inf')
    visited = False
    # constructor
    def __init__(self, label):
        self.label = label
        self.neighbors = []
        self.distances = []
        self.prevNode = None
        self.totalDistance = float('Inf')
        self.visited = False


# find the shortest paths to all nodes recursively
def dijkstra(node):
    # visit all neighbors and update them if necessary
    for i in range(len(node.neighbors)):
        n = node.neighbors[i]
        d = node.distances[i]
        if n.totalDistance > d + node.totalDistance:
            n.prevNode = node
            n.totalDistance = d + node.totalDistance
            heappush(pq, (n.totalDistance, n))
    node.visited = True

    (d, ne) = heappop(pq)
    if not ne.visited:
        dijkstra(ne)

# get the shortest path to the given node
def route(endNode):
    node = endNode
    labels = [endNode.label]
    # distances = []
    while node.label != node.prevNode.label:
        # distances.append(node.totalDistance - node.prevNode.totalDistance)
        node = node.prevNode
        labels.append(node.label)
    labels.reverse()
    return labels
    # distances.reverse()
    # return (labels, distances)

# create a graph
a = node('a')
b = node('b')
c = node('c')
d = node('d')
e = node('e')
f = node('f')
g = node('g')
graph = (a, b, c, d, e, f, g)

# create bidirectional edges of the graph
edges = []
edges.append((b, c, 3))
edges.append((d, e, 1))
edges.append((b, d, 2))
edges.append((c, e, 1))
edges.append((a, b, 1))
edges.append((a, d, 1))
edges.append((c, f, 1))
edges.append((e, f, 3))
edges.append((e, g, 1))

# create adjaceny list of neighbors for each node
for edge in edges:
    edge[0].neighbors.append(edge[1])
    edge[0].distances.append(edge[2])
    edge[1].neighbors.append(edge[0])
    edge[1].distances.append(edge[2])

# print the graph
print 'The graph:'
print
for n in graph:
    print 'Node: ', n.label
    print 'Neighbors:'
    for i in range(len(n.neighbors)):
        print n.neighbors[i].label, n.distances[i]
    print

# find the shortest paths to all neighbors starting w/ the given node
startNode = a
print 'Route start node:', startNode.label
startNode.prevNode = startNode
startNode.totalDistance = 0
dijkstra(startNode)

##print
##print 'The graph after Dijkstra:'
##print
##for n in graph:
##    print 'Node:', n.label
##    print 'totalDistance:', n.totalDistance
##    print 'prevNode:', n.prevNode.label
##    print

# print the shortest path to the given node
endNode = f
print 'Route end node:', endNode.label
print 'Route:', route(endNode)
print 'Total distance:', endNode.totalDistance
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2 回答 2

0

使用pydot，并以不同的方式设置所选边缘的样式。

于 2012-11-22T23:59:47.153 回答
0

显示所有节点，它们之间的距离与它们之间的路径权重成正比。然后用一条有向线连接作为最短路径一部分的每一对。

您可以使用networkx和graphvis在 Python 中创建这样的图表。

于 2012-11-23T00:03:09.123 回答