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For 5 marks, we can get free marks by doing BFS on a grid.
Time Complexity: O(NM)
For another 5 marks, we have to consider cameras.
For each camera, traverse outward in all directions and mark all . squares as invalid, stopping when you hit a wall.
Then when BFSing, you cannot traverse invalid squares. They are essentially walls.
Also be careful: If a camera can see the robot's starting position, it cannot go anywhere.
Time Complexity: O(NM * max(N, M))
For the full solution, we must consider conveyors.
For each conveyor, we can precompute the end points. If the end point is in a camera's path or it ends in a wall, mark the whole conveyor belt as invalid. This means we cannot get on this conveyor belt at all. It is a wall.
Also, if the conveyor belt forms a cycle, it is similarly invalid.
Then when BFSing, when we get on a conveyor, we can 'teleport' to it's endpoint.
Time Complexity: O(NM * max(N, M))