python dijkstra可能存在的问题';s算法的类实现
因此,我一直在尝试用python创建一个图形程序,其中的类实现了dfs、bfs和dijkstra的算法,到目前为止,我已经提出了:python dijkstra可能存在的问题';s算法的类实现,python,python-3.x,Python,Python 3.x,因此,我一直在尝试用python创建一个图形程序,其中的类实现了dfs、bfs和dijkstra的算法,到目前为止,我已经提出了: class Vertex: def __init__(self, name): self.name = name self.connections = {} def addNeighbour(self, neighbour, cost): self.connections[neighbour] = c
class Vertex:
def __init__(self, name):
self.name = name
self.connections = {}
def addNeighbour(self, neighbour, cost):
self.connections[neighbour] = cost
class Graph:
def __init__(self):
self.vertexs = {}
def addVertex(self, newVertex):
new = Vertex(newVertex)
self.vertexs[newVertex] = new
def addEdge(self, src, dest, cost):
self.vertexs[src].addNeighbour(self.vertexs[dest], cost)
def dfs(self, start, end, visited):
visited[start] = True
print(start, end=' ')
if start == end:
# End node found
return True
else:
# Use depth first search
for connection in graph.vertexs[start].connections:
if visited[connection.name] == False:
if self.dfs(connection.name, end, visited) == True:
# Return true to stop extra nodes from being searched
return True
def bfs(self, start, end, visited, queue):
if len(queue) == 0:
# Queue is empty
queue.append(start)
visited[start] = True
currentNode = queue.pop(0)
print(currentNode, end=' ')
if start == end:
# End node found
return True
else:
# Do breadth first search
for connection in graph.vertexs[currentNode].connections:
if visited[connection.name] == False:
# Queue all its unvisited neighbours
queue.append(connection.name)
for newNode in queue:
self.bfs(newNode, end, visited, queue)
def dijkstra(self, current, currentDistance, distances, visited, unvisited):
for neighbour, distance in distances.items():
if neighbour.name not in unvisited:
continue
newDistance = currentDistance + distance
if unvisited[neighbour.name] is None or unvisited[neighbour.name] > newDistance:
unvisited[neighbour.name] = newDistance
visited[current] = currentDistance
del unvisited[current]
if not unvisited:
return True
candidates = [node for node in unvisited.items() if node[1]]
current, currentDistance = sorted(candidates)[0]
self.dijkstra(current, currentDistance, graph.vertexs[current].connections, visited, unvisited)
return visited
def setup():
graphList = {
# Node number: [destination number, cost]
0: {4: 6, 6: 1},
1: {6: 2},
2: {0: 9, 1: 4, 3: 3},
3: {4: 7},
4: {1: 3, 5: 5},
5: {0: 2, 1: 6, 4: 3},
6: {2: 4, 3: 6}
}
graph = Graph()
for i in range(len(graphList)):
graph.addVertex(i)
for dictLength in range(len(graphList)):
for key in list(graphList[dictLength].keys()):
graph.addEdge(dictLength, key, graphList[dictLength][key])
return graph, graphList
graph, graphList = setup()
print("DFS travsersal path from node 1 to node 0:")
graph.dfs(1, 0, [False] * len(graphList))
print()
print("BFS traversal path from node 1 to node 0:")
graph.bfs(1, 0, [False] * len(graphList), [])
print()
print("Shortest possible path from node 1 to 0:")
result = graph.dijkstra(1, 0, graph.vertexs[2].connections, {}, {node: None for node in graphList})
cost = result[len(result) - 1]
path = " ".join([str(arrInt) for arrInt in list(result.keys())])
print(path, "costing", cost)
但我认为输出似乎有问题。如果我想从节点1移动到节点0,则当前输出为:
从节点1到节点0的DFS travsersal路径:
1 6 2 0
从节点1到节点0的BFS遍历路径:
1 6 2 3 0 3 4 5
从节点1到0的最短可能路径:
1034562成本10
但是,我认为输出应该是:
从节点1到节点0的DFS travsersal路径:
1 6 2 0 4 5 3
从节点1到节点0的BFS遍历路径:
1 6 2 3 0 4 5
从节点1到0的最短可能路径:
1 6 2 0成本15
有人看到这有什么问题吗
谢谢 您的代码中实际上有几个问题:
cost=result[len(result)-1]
并不能得到字典中的最后一个元素(字典通常不排序,所以“最后一个元素”甚至不存在!)。您应该将成本检索为cost=result[end]
,其中end
是最后一个节点,在您的示例中为0
result=graph.dijkstra(1,0,graph.vertexs[2].connections,{},{node:None for node in graphList})的形式调用函数
,但是,此函数的第三个参数应该是初始节点的邻居集,因此在您的例子中它应该是graph.vertexs[1].connections
def dijkstra(self, current, currentDistance, distances, visited, unvisited, end):
for neighbour, distance in distances.items():
if neighbour.name not in unvisited:
continue
newDistance = currentDistance + distance
if unvisited[neighbour.name] is None or unvisited[neighbour.name] > newDistance:
unvisited[neighbour.name] = newDistance
visited[current] = currentDistance
if current == end:
return visited
del unvisited[current]
if not unvisited:
return True
candidates = [node for node in unvisited.items() if node[1]]
current, currentDistance = sorted(candidates)[0]
self.dijkstra(current, currentDistance, graph.vertexs[current].connections, visited, unvisited, end)
return visited
print("Shortest possible path from node 1 to 0:")
start = 1
end = 0
result = graph.dijkstra(start, 0, graph.vertexs[start].connections, {}, {node: None for node in graphList}, end)
cost = result[end]
path = " ".join([str(arrInt) for arrInt in list(result.keys())])
print(path, "costing", cost)
并称之为:
def dijkstra(self, current, currentDistance, distances, visited, unvisited, end):
for neighbour, distance in distances.items():
if neighbour.name not in unvisited:
continue
newDistance = currentDistance + distance
if unvisited[neighbour.name] is None or unvisited[neighbour.name] > newDistance:
unvisited[neighbour.name] = newDistance
visited[current] = currentDistance
if current == end:
return visited
del unvisited[current]
if not unvisited:
return True
candidates = [node for node in unvisited.items() if node[1]]
current, currentDistance = sorted(candidates)[0]
self.dijkstra(current, currentDistance, graph.vertexs[current].connections, visited, unvisited, end)
return visited
print("Shortest possible path from node 1 to 0:")
start = 1
end = 0
result = graph.dijkstra(start, 0, graph.vertexs[start].connections, {}, {node: None for node in graphList}, end)
cost = result[end]
path = " ".join([str(arrInt) for arrInt in list(result.keys())])
print(path, "costing", cost)
通过这样做,输出变为
从节点1到0的最短可能路径:
1 6 2 0成本15
非常感谢,现在可以用了。我还注意到bfs()中有一个小问题,它不会停止,因此我使用了与dfs()函数中相同的技术。@Derp Diamonds,如果此解决方案对您有效,并且您接受它,则向上投票并将其标记为接受的答案。