python求带权图的最短路径算法

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# the graph
graph = {}
graph["start"] = {}
graph["start"]["a"] = 6
graph["start"]["b"] = 2

graph["a"] = {}
graph["a"]["fin"] = 1

graph["b"] = {}
graph["b"]["a"] = 3
graph["b"]["fin"] = 5

graph["fin"] = {}

# the costs table
infinity = float("inf")
costs = {}
costs["a"] = 6
costs["b"] = 2
costs["fin"] = infinity

# the parents table
parents = {}
parents["a"] = "start"
parents["b"] = "start"
parents["fin"] = None

processed = []

def find_lowest_cost_node(costs):
lowest_cost = float("inf")
lowest_cost_node = None
# Go through each node.
for node in costs:
cost = costs[node]
# If it's the lowest cost so far and hasn't been processed yet...
if cost < lowest_cost and node not in processed:
# ... set it as the new lowest-cost node.
lowest_cost = cost
lowest_cost_node = node
return lowest_cost_node

# Find the lowest-cost node that you haven't processed yet.
node = find_lowest_cost_node(costs)
# If you've processed all the nodes, this while loop is done.
while node is not None:
cost = costs[node]
# Go through all the neighbors of this node.
neighbors = graph[node]
for n in neighbors.keys():
new_cost = cost + neighbors[n]
# If it's cheaper to get to this neighbor by going through this node...
if costs[n] > new_cost:
# ... update the cost for this node.
costs[n] = new_cost
# This node becomes the new parent for this neighbor.
parents[n] = node
# Mark the node as processed.
processed.append(node)
# Find the next node to process, and loop.
node = find_lowest_cost_node(costs)

print "Cost from the start to each node:"
print costs

为什么终点的父节点初始默认为None
个人觉得 A,B都是终点的父节点

[extend]

这就是图的遍历，
深度遍历或广度遍历都行。
明显只有A是重点父节点：起点-->B-->A-->终点

你图中的数据结构也有问题吧。节点A--B也是一条路径