O（E + V）算法计算给定图上2个节点之间的最短路径数

Question

When Given a graph G with vertices and edges |V| and |E| respectively and vertices u and t, write a O(|E|+|V|) algorithm to compute the number of shortest paths from u to t, ie if there are 5 paths of length 4 with length 4 being the shortest path from u to t then the algorithm will output 5.

I know that the algorithm must somehow incorporate either a DFS or a BFS due to its run-time as each have a O(|E|+|V|) run-time, but I am a bit stuck. I tried implementing something where it would do a DFS repeatedly with the algorithm terminating at t but it this became problematic for deciding which nodes to set as visited and which to reset after each iteraition.

Thanks in advance!

Answer 1

You can use breadth-first search. For each vertex, keep track of:

• the shortest path length from u to that vertex
  • Whenever you process a given vertex, you set this property for all of its neighbors that don't already have this property set.
  • This doubles as a "has already been enqueued" flag: you initially set to a sentinel value such as ɴɪʟ or ∞, and only update it once for any given vertex. So you don't need a separate flag to track visited vertices.
• the number of shortest paths from u to that vertex
  • Whenever you process a given vertex, you increase this property appropriately for all of its neighbors whose shortest path length from u is greater than that of the vertex you're processing.
  • Note that you'll update this property many times for some vertices, but that's OK, because you only update it once per edge.