Let $A$ be a primitive $(0, 1)$-matrix of order $n$, and $\cdots$ then $\text{exp} (A) \leq n + l(n − 2).$ 
Let $A$ be a primitive $(0, 1)$-matrix of order $n$, and let $l$ be the length
  of a shortest cycle in its digraph $G$. Then $$\text{exp} (A) \leq n + l(n − 2).$$ 

Proof: Consider the matrix $A^l$ and its digraph $G^l$. There is an arc in $G^l$ from vertex $v_i$ to vertex $v_j$ exactly when there is a directed $v_i-v_j$ walk of length $l$ in $G$. Let $L$ be the set of vertices of $G^l$ that have loops. Then $L$ has at least $l$ elements, and the loops at vertices of $L$ imply that there is a directed walk in $G^l$ of length $n − 1$ from each vertex in $L$ to each vertex $v_j$. Hence in $G$ there is a directed walk of length $l(n − 1)$ from each vertex in $L$ to each vertex $v_j$. For each vertex $v_i$ there is a directed walk in $G$ from $v_i$ to some vertex in $L$ of length $l_i \leq n − l$, and hence a directed $v_i-v_j$ walk of length $n − l + l(n − 1) = n + l(n − 2)$. Hence $A^{n+l(n−2)}$ is a positive matrix.

Here positive matrix means entry-wise positive.
A primitive matrix is a non-negative matrix A such that $A^k>0$ for some natural integer k.
The exponent of $A$, denoted by $\text{exp} (A)$, is the smallest positive integer $t$ for which $A^t$ is positive.

Having difficulty in understanding: Then $L$ has at least $l$ elements, and the loops at vertices of $L$ imply that there is a directed walk in $G^l$ of length $n − 1$ from each vertex in $L$ to each vertex $v_j$.

Ref: (Encyclopedia of Mathematics and its Applications v. 1) Lowell W. Beineke, Robin J. Wilson, Peter J. Cameron-Topics in Algebraic Graph Theory-Cambridge University Press (2004). Page 64-65.
 A: 
The set $L$ has at least $l$ elements.

If there is a cycle $v_1 \to v_2 \to \dots \to v_l \to v_1$ of length $l$, then each of the vertices $v_1, v_2, \dots, v_l$ has a loop in $G^l$, because there is a walk from $v_i$ back to $v_i$ of length $l$: just go around the cycle. The set $L$ is larger if there is more than one cycle of length $l$.

The loops at vertices of $L$ imply that there is a directed walk in $G^l$ of length $n−1$ from each vertex in $L$ to each vertex $v_j$.

Because $A$ is primitive, an exponent $t$ exists for which all entries of $A^t$ are positive. If all entries of $A^t$ are positive, then all entries of $(A^t)^l = (A^l)^t$ are positive. Going back to graph theory, this means that in $G^l$ there is a walk between any two vertices of length $t$ - so $G^l$ is connected. 
So if $v$ is a vertex in $L$ and $w$ is any vertex, there is a shortest path in $G^l$ from $v$ to $w$. This path has length at most $n-1$, because the most direct way to get from $v$ to $w$ shouldn't visit any vertex more than once. If it has length $n-1$, we're done. If it's shorter than $n-1$, we can get a walk of length $n-1$ from $v$ to $w$ by taking the loop at $v$ some number of times.
