Given a directed graph $D$ of order $n\geq 4$ and a nonempty subset $Y$ of vertices of $D$ such that in $D$ every vertex of $Y$ reachable from every other vertex of $Y$. Assume that for every triple $x,y,z\in Y$ such that $x$ and $y$ are nonadjacent: If there is no arc from $x$ to $z$, then $d(x)+d(y)+d^+(x)+d^-(z)\geq 3n-2$. If there is no arc from $z$ to $x$, then $d(x)+d(y)+d^+(z)+d^-(x)\geq 3n-2$. We prove that there is a directed cycle in $D$ which contains all the vertices of $Y$, except possibly one. This result is best possible in some sense and gives a answer to a question of H. Li, Flandrin and Shu (Discrete Mathematics, 307 (2007) 1291-1297).