1.找出"最便宜"的节点,即可在最短时间内到达的节点
2.更新该节点的邻居和开销,检查是否有前往他们的更短路劲,如果有,就更新其开销
3.重复这个过程,直到对图中的每个节点都这样做
4.计算最短路径
graph = {}
graph["start"] = {}
graph["start"]["a"] = 6
graph["start"]["b"] = 2
# print(graph["start"].keys())
# print(graph["start"]["a"])
graph["a"] = {}
graph["a"]["fin"] = 1
graph["b"] = {}
graph["b"]["a"] = 3
graph["b"]["fin"] = 5
graph["fin"] = {}
infinity = float("inf")
costs = {}
costs["a"] = 6
costs["b"] = 2
costs["fin"] = infinity
parents = {}
parents["a"] = "start"
parents["b"] = "start"
parents["fin"] = None
processed = []
def find_lowest_cost_node(costs):
lowest_cost = float("inf")
lowest_cost_node = None
for node in costs: # 遍历所有的节点
cost = costs[node]
if cost < lowest_cost and node not in processed: # 如果当前节点的开销更低且尚未处理过
lowest_cost = cost # 就将其视为开销最低的节点
lowest_cost_node = node
return lowest_cost_node
node = find_lowest_cost_node(costs) # 在未处理的节点中找出开销最小的节点
while node is not None: # 这个循环在所有节点都被处理过后结束
cost = costs[node]
neighbors = graph[node]
for n in neighbors.keys(): # 遍历当前节点的所有邻居
new_cost = cost + neighbors[n]
if costs[n] > new_cost: # 如果经当前节点前往该邻居更近
costs[n] = new_cost # 就更新该邻居的开销
parents[n] = node # 同时将该邻居的父节点设置为当前节点
processed.append(node) # 将当前节点标记为处理过
node = find_lowest_cost_node(costs) # 找出接下来要处理的节点,并循环
print(graph)
print(costs)
print(parents)