[英]Dijkstra's algorithm Python
給定一個圖和圖中的一個源頂點,找到從源到給定圖中所有頂點的最短路徑。 在這里閱讀更多-> 鏈接
請通過我的代碼 go 並通過指出我的邏輯有什么問題來幫助我。
我的代碼:
from collections import defaultdict
global INT_MAX
INT_MAX = 3 ** 38
class Graph:
def __init__(self, numofVertices):
self.vertList = defaultdict(list)
self.numofVertices = numofVertices
def addEdge(self, u, v, cost):
self.vertList[u].append((v, cost))
self.vertList[v].append((u, cost))
def minDist(self, dist, visited):
for v in range(self.numofVertices):
if dist[v] < INT_MAX and v not in visited:
minIndex = v
return minIndex
def dijsktra(self, src):
dist = [INT_MAX] * self.numofVertices
dist[src] = 0
visited = set()
for _ in range(self.numofVertices):
minVertex = self.minDist(dist, visited)
visited.add(minVertex)
for nbr, edgeCost in self.vertList[minVertex]:
if dist[nbr] > dist[minVertex] + edgeCost and nbr not in visited:
dist[nbr] = dist[minVertex] + edgeCost
return dist
g = Graph(9)
g.addEdge(0, 1, 4)
g.addEdge(0, 7, 8)
g.addEdge(1, 7, 11)
g.addEdge(7, 8, 7)
g.addEdge(7, 6, 1)
g.addEdge(7, 1, 11)
g.addEdge(1, 2, 8)
g.addEdge(2, 3, 7)
g.addEdge(2, 5, 4)
g.addEdge(2, 8, 2)
g.addEdge(6, 8, 6)
g.addEdge(6, 5, 2)
g.addEdge(5, 2, 4)
g.addEdge(5, 3, 14)
g.addEdge(5, 4, 10)
g.addEdge(3, 4, 9)
print(g.dijsktra(0))
當前 Output:
[0, 4, 15, 25, 21, 11, 9, 8, 15] # Index represents the vertex
預計 Output
[0, 4, 12, 19, 21, 11, 9, 8 ,14]
問題出在檢查最小距離的 function 上,我們需要更新當前的最大值,以便我們可以將其與其他未訪問的頂點進行比較,並查看是否存在另一個具有較小值的頂點。
def minDist(self, dist, visited):
max1 = 3 ** 38
minIndex = 0
for v in range(self.numofVertices):
if dist[v] < max1 and v not in visited:
max1 = dist[v]
minIndex = v
return minIndex
minDist
方法有錯誤。 它返回第一個相鄰索引的索引,而不是距離最小的索引。
這個 function 應該如下所示:
def minDist(self, dist, visited):
m = INT_MAX
for v in range(self.numofVertices):
if dist[v] < m and v not in visited:
m = dist[v]
minIndex = v
return minIndex
另外,我不確定您是否必須在 addEdge 中同時分配vertList[u]
和addEdge
vertList[v]
。
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