The correct answer is **B) $5.0$ units**. To find the minimum total cost, we must consider two distinct cases, as the traveler must visit exactly one Service Hub ($C_{2,2}$ or $C_{3,3}$). We will analyze each case separately and then choose the path with the overall minimum cost. **Common Rules and Costs:** * Standard Road cost: $1$ unit. * Express Lane cost: $0.5$ units (if enabled), $1$ unit (if not enabled). * Express Lane enablement: After visiting exactly one Service Hub. * No city revisit. **Case 1: The traveler visits $C_{2,2}$ as the ONLY Service Hub.** This implies the path must avoid $C_{3,3}$. The journey can be logically divided into two phases: **Phase 1: From $C_{1,1}$ to $C_{2,2}$ (before Express Lanes are enabled).** During this phase, all road segments, including those designated as Express Lanes, will cost $1$ unit because no Service Hub has been visited yet. We need to find the shortest path from $C_{1,1}$ to $C_{2,2}$ that avoids $C_{3,3}$. Possible shortest paths (2 segments, $1$ Down, $1$ Right): * $C_{1,1} \rightarrow C_{1,2} \rightarrow C_{2,2}$. Cost: $1+1 = 2$. This path does not visit $C_{3,3}$. * $C_{1,1} \rightarrow C_{2,1} \rightarrow C_{2,2}$. Cost: $1+1 = 2$. This path does not visit $C_{3,3}$. The minimum cost to reach $C_{2,2}$ is $2$ units. **Phase 2: From $C_{2,2}$ to $C_{4,4}$ (Express Lanes are enabled).** Now that $C_{2,2}$ has been visited, Express Lanes cost $0.5$ units. We must find the shortest path from $C_{2,2}$ to $C_{4,4}$ while avoiding $C_{3,3}$ and any previously visited cities (though moving towards $C_{4,4}$ generally avoids revisiting). Let's map out the minimum cost from $C_{2,2}$ to $C_{4,4}$ under these conditions. We can use a modified Dijkstra's algorithm or careful enumeration for this $3 \times 3$ subgrid ($C_{2,2}$ to $C_{4,4}$ with $C_{3,3}$ excluded). Relevant Express Lanes (cost $0.5$): $(C_{2,2}-C_{2,3})$, $(C_{4,2}-C_{4,3})$, $(C_{2,4}-C_{3,4})$. Other Express Lanes are either irrelevant or connect to $C_{3,3}$. Consider the path: $C_{2,2} \xrightarrow{\text{Express: } 0.5} C_{2,3}$ (using $C_{2,2}-C_{2,3}$ Express Lane) $C_{2,3} \xrightarrow{\text{Standard: } 1} C_{2,4}$ $C_{2,4} \xrightarrow{\text{Express: } 0.5} C_{3,4}$ (using $C_{2,4}-C_{3,4}$ Express Lane) $C_{3,4} \xrightarrow{\text{Standard: } 1} C_{4,4}$ Total cost for Phase 2: $0.5 + 1 + 0.5 + 1 = 3$ units. (A path like $C_{2,2} \rightarrow C_{3,2} \rightarrow C_{4,2} \rightarrow C_{4,3} \rightarrow C_{4,4}$ would cost $1+1+0.5+1=3.5$, which is higher.) Total minimum cost for Case 1: $2 \text{ (Phase 1)} + 3 \text{ (Phase 2)} = 5$ units. The complete path could be $C_{1,1} \rightarrow C_{1,2} \rightarrow C_{2,2} \rightarrow C_{2,3} \rightarrow C_{2,4} \rightarrow C_{3,4} \rightarrow C_{4,4}$. All conditions are met: exactly one hub ($C_{2,2}$) visited, no revisits, express lanes used correctly. **Case 2: The traveler visits $C_{3,3}$ as the ONLY Service Hub.** This implies the path must avoid $C_{2,2}$. **Phase 1: From $C_{1,1}$ to $C_{3,3}$ (before Express Lanes are enabled).** All road segments cost $1$ unit. We need the shortest path from $C_{1,1}$ to $C_{3,3}$ that avoids $C_{2,2}$. $C_{1,1}$ to $C_{3,3}$ requires $2$ Down and $2$ Right movements, total $4$ segments. A direct path via $C_{2,2}$ (e.g., $C_{1,1} \rightarrow C_{1,2} \rightarrow C_{2,2} \rightarrow C_{2,3} \rightarrow C_{3,3}$) would cost $4$ but is invalid as it visits $C_{2,2}$. Paths avoiding $C_{2,2}$: * $C_{1,1} \rightarrow C_{1,2} \rightarrow C_{1,3} \rightarrow C_{2,3} \rightarrow C_{3,3}$. Cost: $1+1+1+1 = 4$. (Valid, avoids $C_{2,2}$). * $C_{1,1} \rightarrow C_{2,1} \rightarrow C_{3,1} \rightarrow C_{3,2} \rightarrow C_{3,3}$. Cost: $1+1+1+1 = 4$. (Valid, avoids $C_{2,2}$). The minimum cost to reach $C_{3,3}$ avoiding $C_{2,2}$ is $4$ units. **Phase 2: From $C_{3,3}$ to $C_{4,4}$ (Express Lanes are enabled).** Express Lanes cost $0.5$ units. We need the shortest path from $C_{3,3}$ to $C_{4,4}$ avoiding $C_{2,2}$ and any previously visited cities. $C_{3,3}$ to $C_{4,4}$ requires $1$ Down and $1$ Right movement, total $2$ segments. The Express Lanes potentially useful from $C_{3,3}$ (while heading to $C_{4,4}$ and avoiding $C_{2,2}$) are $(C_{4,2}-C_{4,3})$. However, this would involve going $C_{3,3} \rightarrow C_{3,2} \rightarrow C_{4,2} \rightarrow C_{4,3} \rightarrow C_{4,4}$, which is $0.5+1+0.5+1 = 3$ if we could use $C_{3,3}-C_{3,2}$ as express. But $C_{3,3}-C_{3,2}$ is an express lane from the problem description, so it would cost 0.5. Path cost $0.5 ( ext{E}) + 1 ( ext{S}) + 0.5 ( ext{E}) + 1 ( ext{S}) = 3$. This is longer than direct standard paths. Consider the direct paths: * $C_{3,3} \xrightarrow{\text{Standard: } 1} C_{3,4} \xrightarrow{\text{Standard: } 1} C_{4,4}$. Cost: $1+1 = 2$. * $C_{3,3} \xrightarrow{\text{Standard: } 1} C_{4,3} \xrightarrow{\text{Standard: } 1} C_{4,4}$. Cost: $1+1 = 2$. No Express Lanes offer a shorter path from $C_{3,3}$ to $C_{4,4}$ while avoiding $C_{2,2}$ compared to standard roads. The minimum cost for Phase 2 is $2$ units. Total minimum cost for Case 2: $4 \text{ (Phase 1)} + 2 \text{ (Phase 2)} = 6$ units. **Conclusion:** Comparing the minimum costs from both cases: * Case 1 (via $C_{2,2}$): $5$ units * Case 2 (via $C_{3,3}$): $6$ units The overall minimum cost is $5.0$ units. The final answer is $\boxed{ ext{5.0 units}}$