我对编码比较新，所以对覆盖的，笨拙的代码道歉。
另外f.f.r.这是来自麻省理工学院的开放课程＆＃34;问题集11，最快捷的方式来到麻省理工学院＆＃34;。 （对于以后的问题集，我无法在线找到解决方案..）
我已经实现了Dijkstra算法的大部分成功版本，无论是否有动态编程。在这个实现中，我保留了一个节点列表，以避免循环/无限循环...但是我理解这是我遇到问题的地方。鉴于我的边缘是加权的，在某些情况下我想通过不同的EDGES重新访问相同的NODES，这导致我的代码忽略了正确的解决方案。所以我想找到一种方法来修改我的代码，以便记住哪些边缘已被访问而不是哪个节点...我确实尝试修改它以记住哪些边缘已被访问 - 但我的实现不起作用 - 所以我很乐意看到正确的写作方式。
我的代码失败的示例输入：
# Test case 6
print "---------------"
print "Test case 6:"
print "Find the shortest-path from Building 1 to 32 without going outdoors"
expectedPath6 = ['1', '3', '10', '4', '12', '24', '34', '36', '32']
brutePath6 = bruteForceSearch(digraph, '1', '32', LARGE_DIST, 0)
dfsPath6 = directedDFS(digraph, '1', '32', LARGE_DIST, 0)
print "Expected: ", expectedPath6
print "Brute-force: ", brutePath6
print "DFS: ", dfsPath6
该代码的输出......
---------------
Test case 6:
Find the shortest-path from Building 1 to 32 without going outdoors
Expected: ['1', '3', '10', '4', '12', '24', '34', '36', '32']
Brute-force: ['1', '3', '10', '13', '24', '34', '36', '32']
DFS: ['1', '3', '3', '3', '7', '7', '9', '13', '24', '34', '36', '36', '32']
邻接列表是dict中dict的形式，其中权重存储为元组。要检查强力代码，地图输入或有向图的构建方式，请转到此GitHub here。
def findDist(digraph, start, path):
path = [str(start)] + path
distance = 0
outDistance = 0
for i in range(len(path)-1):
edgeVals = None
listOfDest = digraph.edges[Node(path[i])]
for node in listOfDest:
if node.has_key(Node(path[i+1])):
edgeVals = node.values()
distance = distance + edgeVals[0][0]
outDistance = outDistance + edgeVals[0][1]
break
return (distance, outDistance)
def directedDFS(digraph, start, end, maxTotalDist, maxDistOutdoors, toPrint = False, visited = [], level = 0, memo = {}):
"""
Finds the shortest path from start to end using directed depth-first.
search approach. The total distance travelled on the path must not
exceed maxTotalDist, and the distance spent outdoor on this path must
not exceed maxDisOutdoors.
Parameters:
digraph: instance of class Digraph or its subclass
start, end: start & end building numbers (strings)
maxTotalDist : maximum total distance on a path (integer)
maxDistOutdoors: maximum distance spent outdoors on a path (integer)
Assumes:
start and end are numbers for existing buildings in graph
Returns:
The shortest-path from start to end, represented by
a list of building numbers (in strings), [n_1, n_2, ..., n_k],
where there exists an edge from n_i to n_(i+1) in digraph,
for all 1 <= i < k.
If there exists no path that satisfies maxTotalDist and
maxDistOutdoors constraints, then raises a ValueError.
"""
#TODO
if toPrint:
print start, end
check = start
start = Node(start)
end = Node(end)
if not (digraph.hasNode(start) and digraph.hasNode(end)):
raise ValueError('Start or end not in graph.')
path = [str(start)]
if start == end:
return path
shortestDist = None
shortestOutDist = None
distFilter = (maxTotalDist <= maxDistOutdoors)
for child in digraph.childrenOf(start):
childNode = child.keys()[0]
if (str(childNode) not in visited):
visited = visited + [str(childNode)]
try:
newPath = memo[start, end]
except:
newPath = directedDFS(digraph, str(childNode), str(end), maxTotalDist, maxDistOutdoors, toPrint, visited, level = level + 1)
if newPath == None:
continue
newPathDist = findDist(digraph, start, newPath)
if (distFilter) and (newPathDist[1] <= maxDistOutdoors) and ((shortestDist == None) or (newPathDist[0] < findDist(digraph, start, shortestDist)[0])):
shortestDist = newPath
elif (not distFilter) and ((shortestOutDist == None) or (newPathDist[1] <= findDist(digraph, start, shortestOutDist)[1])):
if (shortestOutDist == None) or (newPathDist[0] <= findDist(digraph, start, shortestOutDist)[0]):
shortestOutDist = newPath
memo[childNode, end] = newPath
if (distFilter) and (shortestDist != None):
path = path + shortestDist
elif (not distFilter) and (shortestOutDist != None):
path = path + shortestOutDist
else:
path = None
if (level == 0) and (not distFilter) and (shortestOutDist == None):
raise ValueError('No such path!')
elif (level == 0) and (not distFilter) and (findDist(digraph, start, shortestOutDist)[1] > maxDistOutdoors):
raise ValueError('No such path!')
elif (level == 0) and (distFilter) and (shortestDist == None):
raise ValueError('No such path!')
elif (level == 0) and (distFilter) and (findDist(digraph, start, shortestDist)[0] > maxTotalDist):
raise ValueError('No such path!')
return path
附加的是预期结果与实际结果的图像 - 忽略后两个已经解决的结果。您可以看到测试用例4和6存在问题。 Code result
谢谢！
