As said elsewhere, this is a problem in computational geometry. The graph structure is of no use.
There is a simple sweepline solution.
Sort the points and polygon vertices by ordinates. Now by a merge-like traversal, you an enumerate all the points in the slab between two successive vertices. By counting the number of edges on the left of every point, you can determine the insideness.
The number of operations will be $O(n\log n+m\log m)$ for the sorts ($n$ points and $m$ edges). Then assuming that the slabs contain on average $e$ edges (a small even number for ordinary polygons), the counting phase will cost like $O(e(n+m))$ operations.
With a little more sophistication, I suspect that this can be limited to $O(n\log m+m)$ in the worst case.