For which families of graphs is Generalized Geography in $P$? As @Marzio mentioned, the following game is known as Generalized Geography.

Given a graph $G=(V,E)$ and a starting vertex $v \in V$, the game is defined as follows:

At each turn (two players alternating), a player chooses $u\in N(v)$, and then the following happens:


$v$, as well as all of its edges, is removed from $G$.
$u\to v$ (i.e. $v$ is updated to be the vertex $u$).


The player who is forced to select a "dead end" (i.e. a vertex with no outgoing edges) loses.

In which graph families is the optimal strategy computable in polynomial time?

For example, it's easy to see that if $G$ is a DAG, then we can easily compute the optimal strategy for the players.
Generalized Geography (GG) is PSPACE-complete even on planar directed bipartite graphs,
but, as reported in:

Hans L. Bodlaender, Complexity of path-forming games, Theoretical Computer Science, Volume 110, Issue 1, 15 March 1993, Pages 215-245

GG (and some other PSPACE-complete variants) are linear-time-solvable in graphs of bounded treewidth.

SIDE NOTE: one of the Generalized Geography variants that has recently been proved to be PSPACE-complete is Tron (Light Cycles game): given an undirected graph, two players pick two different starting vertices, and then take turns, by moving to an adjacent vertex from their respective previous one in each step. The game ends when both players cannot move anymore. The player who traversed more vertices wins (it was conjectured to be PSPACE-complete in 1990 by Bodlaender and Kloks).
Tillmann Miltzow, Tron, a combinatorial Game on abstract Graphs (2011)

Edit: I made a small program to test the game on small $n \times m$ rectangular solid grid graphs (undirected), and the result suggests that it is polynomial time solvable also for this class of graphs (assuming that the first node picked by player A is the top-left node):

               Width n
           1 2 3 4 5 6 7 8 
         1 A B A B A B A B    Winning matrix up to 8x8
         2   B B B B B B B 
         3     A B A B A B 
Height m 4       B B B B B  
         5         A B A B 
         6           B B B 
         7             A B 
         8               B 


Curiously, the same matrix is obtained if player A can choose an arbitrary starting node. 

As said in the comments, I think that the complexity of deciding if there is a winning strategy when GG is played on solid grid graphs (with arbitrary shapes, but without holes) is not known and probably it is not so easy to prove something about it (indeed the - somewhat related - problem of deciding if a solid grid graph has an Hamiltonian path is still open, though deciding if a solid grid graph has an Hamiltonian cycle is polynomial time solvable).

A final trivial note: GG is polynomial time solvable also in complete graphs.