使用 Q 查询遍历图中的最后一个节点

Question

Given a directed graph with N nodes(1<=N<=100000) Each node will have only one outgoing edges but there can be more than one incoming edges. There are Q(1<=Q<=10^5) queries where each query is of 2 types.

1. In first query we have to tell if we start traversing the graph from node 'A' then which is the last node where we stop. If we never stop then return -1.
2. second type of query is that we can delete an outgoing edges of node 'A'

I know we can solve this ques in O(N) complexity per query(with overall complexity of QN ), but since the no of queries are high (10^5) this doesn't seems to be the efficient solution?

Any idea how to solve this with better time complexity? Thanks

Answer 1

If you don't need to answer queries online, the easy way is to implement union-find with path compression and process the input backward. Initialize by linking each arc that is never deleted (with the special case below). Scanning backward, for a query of the second type, add the arc, unless it would create a cycle, in which case link the tail to a special vertex with id -1. To answer the first kind of query, find the root with path compression. The running time will be O((Q + N) log N).