Subtask 1

Since node $A$ is equal to $X$, Andy can go straight to $Y$. Thus, the answer is the shortest distance from $X$ to $Y$.

Time Complexity: $\mathcal{O}(M\log N)$

Subtask 2

Let's call the shortest distance between $U$ and $V$ to be $d_{u,v}$.

For any point $A$, there must exist a point $Z$ such that the shortest path goes $X\to Z\to A\to Z\to Y$. This path results in a cost of $d_{X,Z}+d_{Z,A}+d_{Z,Y}$ ($d_{A,Z}$ is ignored because all nodes on the path have already been unlocked when traversing from $Z$ to $A$).

Building on that, let's analyze $d_{X,Z}+d_{Z,A}+d_{Z,Y}$.

$d_{X,Z}+d_{Z,A}+d_{Z,Y}$ can be separated into two parts, $d_{X,Z}+d_{Z,Y}$ and $d_{Z,A}$. $d_{X,Z}+d_{Z,Y}$ doesn't depend on $A$, and can be precomputed by running shortest path twice from nodes $X$ and $Y$. For every query, run shortest path from node $A$ and determine the minimum value of $d_{X,Z}+d_{Z,A}+d_{Z,Y}$ for all possible node $Z$.

Time Complexity: $\mathcal{O}(QM\log N)$

Subtask 3

Set up an additional node $O$, and build new edges with every single node in the graph. The edge between $O$ and any node $U$ is going to have a weight of $d_{X,U}+d_{U,Y}$. Run shortest path algorithm from node $O$, and the shortest path from node $O$ to any node $U$ is the answer for when $A=U$. Thus, each query can be answered in $\mathcal{O}(1)$.

Time Complexity: $\mathcal{O}(M\log N+Q)$