查找 2 个树节点之间的路径
[英]Finding path between 2 tree nodes
问题描述
how can i print the path between any 2 query nodes in a tree not necessary a binary tree?
如何打印树中任意 2 个查询节点之间的路径，而不需要二叉树？ I have been using dfs and storing the paths in a vector for each query and printing it.我一直在使用 dfs 并将路径存储在每个查询的向量中并打印出来。 But if input query no.但是如果输入查询号。 is too large q<=10^5,my algorithm which is of o(n q)(may be not sure) complexity fails.n=no of nodes in the tree.Can anyone help me with some better optimization so that the time complexity reduces may be o(n logq) or o(q*logn).n<=10^5.If any precomputation required suggest me so. q<=10^5 太大，我的算法复杂度为 o(n q)（可能不确定）失败。n=树中没有节点。谁能帮我做一些更好的优化，以便时间复杂性降低可能是 o(n logq) 或 o(q*logn).n<=10^5。如果需要任何预计算建议我这样做。
1 个解决方案
解决方案1
0 2020-04-07 13:53:50
Have you looked at using either the UCS or A* algorithms?您是否考虑过使用 UCS 或 A* 算法？ A* with a good heuristic will allow you to visit the least amount of nodes具有良好启发式的 A* 将允许您访问最少数量的节点
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