如果您嘗試在所有節點上循環，可以對初始值current執行循環。這將需要對代碼進行最少的修改：nodes = ('A', 'B', 'C', 'D', 'E', 'F', 'G')

distances = {

'B': {'A': 5, 'D': 1, 'G': 2},

'A': {'B': 5, 'D': 3, 'E': 12, 'F' :5},

'D': {'B': 1, 'G': 1, 'E': 1, 'A': 3},

'G': {'B': 2, 'D': 1, 'C': 2},

'C': {'G': 2, 'E': 1, 'F': 16},

'E': {'A': 12, 'D': 1, 'C': 1, 'F': 2},

'F': {'A': 5, 'E': 2, 'C': 16}}

for start in nodes:

current = start

currentDistance = 0

unvisited = {node: None for node in nodes}

visited = {}

unvisited[current] = currentDistance

while True:

for neighbour, distance in distances[current].items():

if neighbour not in unvisited: continue

newDistance = currentDistance + distance

if unvisited[neighbour] is None or unvisited[neighbour] > newDistance:

unvisited[neighbour] = newDistance

visited[current] = currentDistance

del unvisited[current]

if not unvisited: break

candidates = [node for node in unvisited.items() if node[1]]

current, currentDistance = sorted(candidates, key = lambda x: x[1])[0]

print(' Shortest distances from %s ' % start)

print(visited)

基本上，我對start進行了一個循環，并將初始的current設置為start。我還在末尾添加了一個打印輸出，告訴您顯示信息的起始節點。在

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