更新感谢一些社区成员的评论,我意识到有一些类似的问题,但它们可能有点不同,请允许我进一步解释。实际上,我希望在实际问题中使用相同的方法,所以简单地说:
完全允许在不同路径中重复使用边
从a到b的唯一(或新)路径定义为具有任何不同顶点的顶点集合。
让我使用一个来自python数据结构和算法分析的测验,由bradley.n miller和david l。让我来解释我的问题。
答:考虑把单词傻瓜转化为圣人的任务,也称为单词梯形问题。在解决单词阶梯问题时,一次只能替换一个字母,并且每个步骤的结果必须是一个单词,而不是不存在的单词。
输入:
FOUL
FOOL
FOIL
FAIL
COOL
FALL
POOL
PALL
POLL
POLE
PALE
PAGE
SALE
POPE
POPE
SAGE
我们可以很容易地找到从傻瓜到圣人的道路,正如布拉德利所示:在这里输入图像描述
我使用广度优先搜索(bfs)来解决问题:
class Vertex:
def __init__(self, key, value = None):
self.id = key
self.connectedTo = {}
self.color = 'white'
self.dist = sys.maxsize
self.pred = []
self.disc = 0
self.fin = 0
self.value = value,
#self.GraphBulided = False
self.traverseIndex = 0
self.predNum = 0
def addNeighbor(self, nbr, weight=0):
self.connectedTo[nbr] = weight
def __str__(self):
return '{} connectedTo: {}'.format(self.id, \
str([x.id for x in self.connectedTo]))
def setColor(self, color):
self.color = color
def setDistance(self, d):
self.dist = d
# I want store all Pred for next traverse so I use a list to do it
def setPred(self, p, list = False):
if not list:
self.pred = p
else:
self.pred.append(p)
self.predNum += 1
def setDiscovery(self,dtime):
self.disc = dtime
def setFinish(self,ftime):
self.fin = ftime
# def setGraphBulided(self, tag = True):
# self.GraphBulided = tag
def getFinish(self):
return self.fin
def getDiscovery(self):
return self.disc
def getPred(self):
if isinstance(self.pred, list):
if self.traverseIndex < self.predNum:
return self.pred[self.traverseIndex]
else:
return self.pred[-1]
else:
return self.pred
def __hash__(self):
return hash(self.id)
def getPredById(self):
if self.traverseIndex < self.predNum and isinstance(self.pred, list):
pred = self.pred[self.traverseIndex]
self.traverseIndex += 1
print("vertix {}: {} of {} preds".format(self.id, self.traverseIndex, self.predNum))
return [pred, self.traverseIndex]
else:
pred = None
return [pred, None]
def getCurrPredStaus(self):
#if not self.pred:
# return None
return self.predNum - self.traverseIndex
def getDistance(self):
return self.dist
def getColor(self):
return self.color
def getConnections(self):
return self.connectedTo.keys()
def getId(self):
return self.id
def getWeight(self, nbr):
return self.connectedTo[nbr]
def getValue(self):
return self.value
def findPath(self, dest):
pass
class Graph:
def __init__(self):
self.vertList = {}
self.numVertics = 0
self.verticsInSerach = set()
self.GraphBulided = False
def addVertex(self, key, value = None):
self.numVertics = self.numVertics + 1
newVertex = Vertex(key, value=value)
self.vertList[key] = newVertex
return newVertex
def getVertex(self, n):
if n in self.vertList:
return self.vertList[n]
else:
return None
def __contains__(self, n):
return n in self.vertList
def addEdge(self, f, t, cost = 0, fvalue = None, tvalue = None):
if f not in self.vertList:
nv = self.addVertex(f, fvalue)
if t not in self.vertList:
nv = self.addVertex(t, tvalue)
self.vertList[f].addNeighbor(self.vertList[t], cost)
def setGraphBulided(self, tag = True):
self.GraphBulided = tag
def getVertices(self):
return self.vertList.keys()
def setGraphBulided(self, tag = True):
self.GraphBulided = tag
def setSerachedVertixs(self, vertix):
self.verticsInSerach.add(vertix)
def getGraphBulided(self):
return self.GraphBulided
def getSerachedVertixs(self):
return self.verticsInSerach
def __iter__(self):
return iter(self.vertList.values())
def __hash__(self):
hashIds = [x for x in self.getVertices()]
if len(hashIds) > 0 and hashIds[0]:
return hash(', '.join(hashIds))
else:
return None
下面是一些用于构建图形的附加函数
def buildGraph(wordFile, DFSgraph = False):
d = {}
g = Graph()
if DFSgraph:
g = DFSGraph()
wfile = open(wordFile)
for line in wfile:
word = line[:-1]
for i in range(len(word)):
bucket = word[:i] + '_' + word[i+1:]
if bucket in d:
d[bucket].append(word)
else:
d[bucket] = [word]
for bucket in d.keys():
for word1 in d[bucket]:
for word2 in d[bucket]:
if word1 != word2:
g.addEdge(word1, word2)
wfile.close()
return g
class Queue:
def __init__(self):
self.items = []
def isEmpty(self):
return self.items == []
def enqueue(self, item):
self.items.insert(0,item)
def dequeue(self):
return self.items.pop()
def size(self):
return len(self.items)
def bfs(g, start, listpred = False):
start.setDistance(0)
start.setPred(None)
vertQueue = Queue()
vertQueue.enqueue(start)
while (vertQueue.size() > 0):
currentVert = vertQueue.dequeue()
if currentVert.getConnections():
g.setSerachedVertixs(currentVert)
for nbr in currentVert.getConnections():
#print('sreach {}'.format(currentVert.getId()))
if (nbr.getColor() == 'white' or nbr.getColor() == 'gray'):
nbr.setColor('gray')
nbr.setDistance(currentVert.getDistance() + 1)
if nbr.predNum > 0 and currentVert.getId() not in [x.getId() for x in nbr.pred]:
nbr.setPred(currentVert, listpred)
elif nbr.predNum == 0:
nbr.setPred(currentVert, listpred)
vertQueue.enqueue(nbr)
currentVert.setColor('black')
因此,我们可以很容易地找到所需的最短路径(如果我们只为一个vertix存储一个pred)。
wordGraph = buildGraph('fourletterwords1.txt', DFSgraph=False)
bfs(wordGraph, wordGraph.getVertex('FOOL'), listpred=True)
def traverse(y):
x=y
while(x.getPred()):
print(x.getPred())
x = x.getPred()
print(x.getId())
traverse(wordGraph.getVertex('SAGE'))
但是,我仍然不知道如何正确地追踪所有路径,你能给我一些建议吗?
1条答案
按热度按时间bq8i3lrv1#