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Jan 28, 2015 at 10:11 comment added Chet Any ideas how you'd do this if you want to change an edge capacity?
Feb 20, 2012 at 13:02 vote accept Saeed
Feb 18, 2012 at 15:40 history edited Dmytro Korduban CC BY-SA 3.0
for the sake of homogeneity
Feb 13, 2012 at 15:56 comment added Dmytro Korduban When you push 1 unit of flow from $v$ to $u$, you decrease $c_f(v,u)$ by 1 and increase $c_f(u,v)$ by 1 because the flow is anti-symmetric ($f(v,u) = -f(u,v)$). This defines true residual graph, so everything works well in it.
Feb 13, 2012 at 15:49 comment added Saeed As I understand in removing node, By $\tilde{f^v}$ you will calculate flow in residual graph, but I think it's not true, in fact in residual graph you have some edges which used in calculating $f^v$ and you should add extra capacity to this edges, then calculating $\tilde{f^v}$, then using $\Delta$.
Feb 13, 2012 at 15:44 comment added Dmytro Korduban After reading last vzn's answer I've found the similar approach described at the page 90 of this.
Feb 13, 2012 at 14:58 history answered Dmytro Korduban CC BY-SA 3.0