network flow sight fix

src/routes/note/network-flow/+page.md

@@ -554,9 +554,9 @@ Dinic's Algorithm can terminate in $O(\sqrt{|V|}|E|)$ time for bipartite matchin
 
 Since $c$ is constant $1$ for all edges, a visit to an edge either clear the capacity or find it has capacity $0$. Therefore, in a DFS between two BFS operations, the number of edges visited is at most $|E|$. Therefore, the time complexity for each DFS is $O(|E|)$.
 
-Meanwhile, the flow in arbitrary iteration is integer-valued, it gives that the flow can be divide into several vertices-disjoint paths except the source $s$ and sink $t$ because if two paths share a vertex, there every vertices either only an in-degree (from the source $s$) or only an out-degree (to the sink $t$) with capacity $1$ and two paths violate the capacity constraint.
+Meanwhile, the flow in arbitrary iteration is integer-valued, it gives that the flow can be divide into several vertices-disjoint paths except the source $s$ and sink $t$ because if two paths share a vertex, there every vertices either only an in-degree (from the source $s$) or only an out-degree (to the sink $t$) with capacity $1$ and two paths violate the capacity constraint. Notice that this holds during the whole process, because when a flow pass through a vertex, it does not change the in or out degree of the vertex. And no matter how the flow is argumented, the number of argumenting paths (even not vertices-disjoint) is same because the capacity of the edge is $1$. That mean we can calculate the number of vertices-disjoint paths other than counting the number of argumenting paths Dinic's Algorithm found.
 
-Consider we have relabeled the vertices $\sqrt{|V|}$ times, the number of vertices-disjoint paths is at least $\sqrt{|V|}$. According to the strictly increasing level of the sink $t$, the length of newly-found path is at least $\sqrt{|V|}$. Therefore, there are no more than $|V| / \sqrt{|V|} = \sqrt{|V|}$ extra vertices-disjoint paths to be found. Since number of iteration is no more than this, the total number of iteration is the $\sqrt{|V|}$ done iteration plus the $\sqrt{|V|}$ extra iteration, which is $O(\sqrt{|V|})$.
+Consider we have relabeled the vertices $\sqrt{|V|}$ times, the number of vertices-disjoint paths is at least $\sqrt{|V|}$. According to the strictly increasing level of the sink $t$, the length of newly-found path is at least $\sqrt{|V|}$. Therefore, there are no more than $|V| / \sqrt{|V|} = \sqrt{|V|}$ extra vertices-disjoint paths to be found in current residual network $G_f$, or in another word, no more than $\sqrt{|V|}$ argumenting paths on current $G_f$. Since number of iteration is no more than this, the total number of iteration is the $\sqrt{|V|}$ done iteration plus the $\sqrt{|V|}$ extra iteration, which is $O(\sqrt{|V|})$.
 
 Therefore, the total time complexity is $O(\sqrt{|V|}|E|)$.