python解决最短路径问题：Floyd-Warshall算法

昨晚做的华为实习生笔试题第三题的解答就涉及到最短路径问题，今天查阅资料l，重新做了一下。

主要思路：

1.根据天气状况更新路线图hmap

2.根据最新的路线图hmap，运用最短路径算法Floyd-Warshall算法,求得任意两城市之间最短路径所必须经过的城市，放在path矩阵中（A矩阵存放对应的权值）

3.然后编写一个函数getRoutes(path,start,end)，根据指定的起始点和终点获得路径。

代码：

#coding= utf-8
#最短路径问题：能到X的所有城市（除去自身），找到时间最小对应的城市M，然后找5到M的时间最小对应的城市，不断递推
#大雾城市：不可从该城市出发，也不可到达该城市，对应行列时间为1000
#从城市5出发到达任意城市的最短时间对应的路径，及

#根据大雾城市设置新的map,y中存放的为大雾城市列表
from collections import defaultdict
from heapq import *
'''
def dijkstra_raw(edges, from_node, to_node):
g = defaultdict(list)
for l,r,c in edges:
g[l].append((c,r))
q, seen = [(0,from_node,())], set()
while q:
(cost,v1,path) = heappop(q)
if v1 not in seen:

4000
seen.add(v1)
path = (v1, path)
if v1 == to_node:
return cost,path
for c, v2 in g.get(v1, ()):
if v2 not in seen:
heappush(q, (cost+c, v2, path))
return float("inf"),[]

def dijkstra(edges, from_node, to_node):
len_shortest_path = -1
ret_path=[]
length,path_queue = dijkstra_raw(edges, from_node, to_node)
if len(path_queue)>0:
len_shortest_path = length ## 1. Get the length firstly;
## 2. Decompose the path_queue, to get the passing nodes in the shortest path.
left = path_queue[0]
ret_path.append(left) ## 2.1 Record the destination node firstly;
right = path_queue[1]
while len(right)>0:
left = right[0]
ret_path.append(left) ## 2.2 Record other nodes, till the source-node.
right = right[1]
ret_path.reverse() ## 3. Reverse the list finally, to make it be normal sequence.
return len_shortest_path,ret_path
'''
def setNewMap(y):
t = [[0,2,10,5,3,1000],
[1000,0,12,1000,1000,10],
[1000,1000,0,1000,7,1000],
[2,1000,1000,0,2,1000],
[4,1000,1000,1,0,1000],
[3,1000,1,1000,2,0]]
if len(y) == 1 and y[0] == '0':
return t
#字符串变为int
for i in xrange(len(y)):
y[i] = int(y[i])
#对应城市的行变为1000
for i in xrange(len(y)):
t[y[i]-1] = [1000,1000,1000,1000,1000,1000]
#对应城市的列变为1000
for i in xrange(6):
for j in y:
t[i][j-1] = 1000
return t

#根据最短路径矩阵，输出两点之间对应的路线
def getRoutes(path,start,end): #最短路径矩阵
rts = [] #存放路线
rts.append(start)
i = start-1
while path[i][end-1] != -1:
rts.append(path[i][end-1]+1)
i = path[i][end-1]
rts.append(end)
return rts
#求最短路径:指定起点sn,终点en和地图tmp
def shortestPath(tmp):
A = tmp
path= [] #存放到某点必须经过的路径点
for i in xrange(6):
path.append([-1,-1,-1,-1,-1,-1])
for k in xrange(6):
for i in xrange(6):
for j in xrange(6):
if A[i][j] > A[i][k] + A[k][j]:
A[i][j] = A[i][k] + A[k][j]
path[i][j] = k
print 'A:%s'%(A)
print 'path:%s'%(path)
return A,path

hmap = [[0,2,10,5,3,1000],
[1000,0,12,1000,1000,10],
[1000,1000,0,1000,7,1000],
[2,1000,1000,0,2,1000],
[4,1000,1000,1,0,1000],
[3,1000,1,1000,2,0]]
'''
lines = [] #存放最短路径
#目的城市
X = raw_input()
#大雾城市，多个大雾城市时以,分割
Y = []
while True:
ins = raw_input()
if not ins:
break
Y = ins.split(',')

#更新当前路径地图
tmp = setNewMap(Y)
'''
Y = [4]
tmp = setNewMap(Y)
a,path = shortestPath(tmp)
rts = getRoutes(path,5,2)
print rts运行结果：



这一个测试用例通过了。

总结：

分析问题，剖析出对应的算法，稍加改造运行实现。

最短路径算法参考链接：

http://www.cnblogs.com/biyeymyhjob/archive/2012/07/31/2615833.html