Obviously, the route of driving is a non-self-intersecting closed polyline. Regard each crossing as a vertex, then the route is regarded as a 100-gon.An interior angle may be greater than or equal to a straight angle.. By the formula of the sum of the angles of the polygon, the sum of all interior angles is $98 \times 180^{\circ}$. Note that the interior angle can only be $90^{\circ}, 180^{\circ}$ or $270^{\circ}$, if there are $a$ angles of $90^{\circ}, b$ angles of $270^{\circ}$, then $90 a+270 b+180(100-a-b)=$ $98 \times 180$, so $a-b=4$. If Bob drives clockwise, then $90^{\circ}, 180^{\circ}$ and $270^{\circ}$ corrsponds to turn right, go straight and turn left, respectively. The total time on waiting at the crossings is $(100-a-b)+2 b=100-(a-b)=96($ min $)$; If Bob drives clockwise, then $90^{\circ}, 180^{\circ}$ and $270^{\circ}$ corrsponds to turn left, go straight and turn right, respectively. The total time on waiting at the crossings is $(100-a-b)+2 a=100+(a-b)=104(\min )$. Therefore, $S=96$, and (C) is correct. Note: If we ignore the waiting time on the beginning/ending crossing, the total time on waiting can be decreased by 2 minutes (Bob can choose a left-turn crossing as the beginning), we have that $S=94$, but do not affect the correct choice.