$G = (V, A)$ is a directed graph where all arcs in $A$ have a capacity of $1$. The shortest length path in $G$ from $s$ to $t$ consists of exactly $d$ arcs. G has a total of $m$ arcs and $n$ vertices.
(a) Prove that the maximum $s-t$ flow has value $O\left(\frac{m}{d}\right)$.
(b) Prove that the minimum capacity $s-t$ cut has value $O\left(\frac{n^2}{d^2}\right)$.
For part a), I was thinking that since there are going to be $m$ arcs total, the value would be in the order of $m$. I am thinking that by diving this $m$ by $d$, we get the total number of arcs used when gathering the maximum $s-t$ flow through the Ford-Fulkerson algorithm. But I am not entirely sure how to prove this or if this is actually the case.
For part b), I am also quite confused as I am not sure how I should go about finding either of the numerator or denominator. I was thinking about splitting the graph in two to represent the cut but that has not been helping me to compute the value.
I was also wondering that since all arcs have a capacity of $1$, wouldn't the value of the maximum flow and the minimum cut both just be $1$? So do I have to show that both $\frac{m}{d}$ and $\frac{n^2}{d^2}$ are equal to $1$?
I would greatly appreciate any help!