Stackelberg Max Closure with Multiple Followers

In a Stackelberg max closure game, we are given digraph whose vertices correspond to projects from which firms can choose and arcs represent precedence constraints. Some under the control of leader who sets prices in first stage while second stage, feasible subset maximum value. For single follower, leader’s problem finding revenue-maximizing be solved strongly polynomial time. this paper, focus on setting with multiple followers distinguish two situations. case only one copy each project is available (limited supply), show that two-follower solvable time, whereas three or more NP-hard. unlimited supply, is, when sufficient copies available, already APX-hard. As side result, prove min vertex cover bipartite graphs follower