练习 Dijkstra 最短路径算法。
#coding: utf-8
# Author: woodfox, Oct 14, 2014
# http://en.wikipedia.org/wiki/Dijkstra%27s_algorithm
"""
Let the node at which we are starting be called the initial node. Let the distance of node Y be the distance from the initial node to Y. Dijkstra's algorithm will assign some initial distance values and will try to improve them step by step.
1. Assign to every node a tentative distance value: set it to zero for our initial node and to infinity for all other nodes.
2. Mark all nodes unvisited. Set the initial node as current. Create a set of the unvisited nodes called the unvisited set consisting of all the nodes.
3. For the current node, consider all of its unvisited neighbors and calculate their tentative distances. Compare the newly calculated tentative distance to the current assigned value and assign the smaller one. For example, if the current node A is marked with a distance of 6, and the edge connecting it with a neighbor B has length 2, then the distance to B (through A) will be 6 + 2 = 8. If B was previously marked with a distance greater than 8 then change it to 8. Otherwise, keep the current value.
4. When we are done considering all of the neighbors of the current node, mark the current node as visited and remove it from the unvisited set. A visited node will never be checked again.
5. If the destination node has been marked visited (when planning a route between two specific nodes) or if the smallest tentative distance among the nodes in the unvisited set is infinity (when planning a complete traversal; occurs when there is no connection between the initial node and remaining unvisited nodes), then stop. The algorithm has finished.
6. Select the unvisited node that is marked with the smallest tentative distance, and set it as the new "current node" then go back to step 3.
"""
vertices = [1, 2, 3, 4, 5, 6]
# element structure is: [node1, node2, cost]
edges = [[1, 6, 14], [1, 2, 7], [1, 3, 9], [3, 6, 2], [2, 3, 10], [3, 4, 11], [2, 4, 15], [6, 5, 9], [5, 4, 6]]
def get_neighbors(node):
# key is neighbor's node number, value is cost
result = {}
for edge in edges:
if edge[0] == node:
other = edge[1]
elif edge[1] == node:
other = edge[0]
else:
continue
result[other] = edge[2] # cost
return result
#print(get_neighbors(3))
#print(get_neighbors(5))
MAX_VALUE = 10000
def shorted_path(start, end):
unvisited = []
visited = []
costs = {}
#result_path = []
for v in vertices:
if v == start:
costs[v] = 0
else:
costs[v] = MAX_VALUE
unvisited.append(v)
#print unvisited
#print costs
current = start
#result_path.append(current)
while True:
neighbors = get_neighbors(current)
min_cost_of_neighbor = MAX_VALUE
min_cost_n = None
if len(neighbors.keys()) > 0:
for n, cost in neighbors.iteritems():
if n in visited:
continue
new_cost = costs[current] + cost
if new_cost < costs[n]:
costs[n] = new_cost
if costs[n] < min_cost_of_neighbor:
min_cost_of_neighbor = costs[n]
min_cost_n = n
visited.append(current)
#print visited, unvisited, current, min_cost_n
unvisited.remove(current)
if min_cost_n != None:
current = min_cost_n
if end in visited:
break
#print costs
return costs[end]
print shorted_path(1, 5)
测试：
Running “dijkstra_test.py”…
Python 2.7.8teaser
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Program exited with code #0 after 0.05 seconds.