问题
We are given a weighed graph G and its Shortest path distance's matrix delta. So that delta(i,j) denotes the weight of shortest path from i to j (i and j are two vertexes of the graph). delta is initially given containing the value of the shortest paths. Suddenly weight of edge E is decreased from W to W'. How to update delta(i,j) in O(n^2)? (n=number of vertexes of graph) The problem is NOT computing all-pair shortest paths again which has the best O(n^3) complexity. the problem is UPDATING delta, so that we won't need to re-compute all-pair shortest paths.
More clarified : All we have is a graph and its delta matrix. delta matrix contains just value of the shortest path. now we want to update delta matrix according to a change in graph: decreased edge weight. how to update it in O(n^2)?
回答1:
If edge E from node a to node b has its weight decreased, then we can update the shortest path length from node i to node j in constant time. The new shortest path from i to j is either the same as the old one or it contains the edge from a to b. If it contains the edge from a to b, then its length is delta(i, a) + edge(a,b) + delta(b, j).
From this the O(n^2) algorithm to update the entire matrix is trivial, as is the one dealing with undirected graphs.
回答2:
http://dl.acm.org/citation.cfm?doid=1039488.1039492 http://dl.acm.org.ezp.lib.unimelb.edu.au/citation.cfm?doid=1039488.1039492
Although they both consider increase and decrease. Increase would make it harder. On the first one, though, page 973, section 3 they explain how to do a decrease-only in n*n.
And no, the dynamic all pair shortest paths can be done in less than nnn. wikipedia is not up to date I guess ;)
回答3:
Read up on Dijkstra's algorithm. It's how you do these shortest-path problems, and runs in less than O(n^2) anyway.
EDIT There are some subtleties here. It sounds like you're provided the shortest path from any i to any j in the graph, and it sounds like you need to update the whole matrix. Iterating over this matrix is n^2, because the matrix is every node by every other, or n*n or n^2. Simply re-running Dijkstra's algorithm for every entry in the delta matrix will not change this performance class, since n^2 is greater than Dijkstra's O(|E|+|V|log|V|) performance. Am I reading this properly, or am I misremembering big-O?
EDIT EDIT It looks like I am misremembering big-O. Iterating over the matrix would be n^2, and Dijkstra's on each would be an additional overhead. I don't see how to do this in the general case without figuring out exactly which paths W' is included in... this seems to imply that each pair should be checked. So you either need to update each pair in constant time, or avoid checking significant parts of the array.
来源：https://stackoverflow.com/questions/4467884/updating-shortest-path-distances-matrix-if-one-edge-weight-is-decreased