network flow part 4

src/lib/component/content/Proof.svelte

@@ -17,7 +17,7 @@
 	<span class="child">
 		<slot />
 	</span>
-	<span class="qed">⯀</span>
+	<span class="qed">■</span>
 </div>
 
 <style>

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

@@ -549,7 +549,17 @@ Dinic's Algorithm can terminate in $O(\sqrt{|V|}|E|)$ time for bipartite matchin
 
 </State>
 
-To be continued.
+<Proof>
+
+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.
+
+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 $O(\sqrt{|V|})$.
+
+Therefore, the total time complexity is $O(\sqrt{|V|}|E|)$.
+
+</Proof>
 
 ## Min-Cost Flow