问题描述
我知道使用邻接矩阵的 DFS 的时间复杂度是 O(V+E),假设顶点是 V,边是 E。但是我如何从这段代码中得到这个复杂度?
from task1 import graph
out = open("output3.txt","w")
visited = []
def dfs_visit(graph,s):
for v in graph[s]:
if v not in visited:
visited.append(v)
dfs_visit(graph,v)
def dfs(graph,end):
for v in [*graph]:
if v not in visited:
visited.append(v)
dfs_visit(graph,v)
out.write("Places: ")
# This part can be skipped as its for printing the values
for v in visited:
if v == end:
out.write(v)
break
out.write(v+" ")
return
# Tester Codes
end = list(graph.keys())[-1]
dfs(graph,end)
解决方法
暂无找到可以解决该程序问题的有效方法,小编努力寻找整理中!
如果你已经找到好的解决方法,欢迎将解决方案带上本链接一起发送给小编。
小编邮箱:dio#foxmail.com (将#修改为@)