Interesting problem on tree
For a fixed root, there are two kinds of path: the ones which pass the root and the ones which don't. The first kind is special since we can seperate a path $(a,b)$ into $(a,r)$ and $(r,b)$, which might be easier to deal with via proper preprocessing. Consider following divide and conquer procedure:

Solve(T):
  Fix a root r
  for each a
    for each b s.t. (a,b) pass r
      AR[a]+=calc(a,b)
  for each branch T_i of r (which is also a tree)
    Solve(T_i)


Note that we neither miss a pair nor count a pair twice since $(a,b)$ don't pass $r$ if and only if they are in the same branch. In addition, it's not hard to prove that trees in depth-$d$ recursive call don't overlap. Therefore, if we can do the non-recursive part in $O(\text{size of tree})$, the complexity would be $O(\text{recursive depth}\times n)$. If we always choose $r$ as the centroid, recursive depth is bounded by $O(\log n)$ and hence the complexity would be $O(n \log n)$. Now let's see how to solve the non-recursive part.

For a pair of node $(a,b)$, let $\text{path}(a,b)$ denote the set of distinct values on the path from $a$ to $b$, and $I_{ab}(v)$ denote the indicator function which is $1$ when $v\in \text{path}(a,b)$ and $0$ otherwise. It's not hard to see $\text{calc}(a,b)=|\text{path}(a,b)|=\sum_vI_v$.

Recall that in each phase, we choose the centroid $r$ as the root and consider only the paths passing the root. Let $B_a=\{b\in T:(a,b)\text{ pass }r\}$. Using the indicator function above, we have $\sum_{b\in B_a}\text{calc}(a,b)=\sum_{b\in B_a}\sum_v I_{ab}(v)=\sum_v\sum_{b\in B_a} I_{ab}(v)$. Therefore, computing $\sum_{b\in B_a} I_{ab}(v)$ for each $a$ and $v$ also gives us the answer. (You can interpret this as the number of $b$ such that $v\in \text{path}(a,b)$.) To do this, consider the set $S$ for each value $v$ such that $S_T(v)=\{u\in T: v\in\text{path}(r,u)\}$. We need to maintain $|S(v)|$ for the whole tree and similarly $|S_i(v)|$ for the $i$-th branch of $r$. We will see how to implement this later.

Since $B_r=T$, $\sum_{b\in B_r}\text{calc}(r,b)$ is exactly $\sum_v|S(v)|$. Next, consider node $a$ in the $i$-th branch. Let $r_i$ denote the $i$-th child of $r$. For each node $b$ which is not in the $i$-th branch, we can split path $(a,b)$ into $(r,b)$ and $(r_i,a)$. Therefore, $$\sum_{b\in B_a}\text{calc}(a,b)=\sum_{b\in B_a}\text{calc}(r,b)+\sum_{b\in B_a}|\text{path}(r_i,a)\backslash\text{path}(r,b)|$$
It's not hard to see that the first term equals to $\sum_v(|S_T(v)|-|S_i(v)|)$, so let's focus on the second term. Now consider each node $u$ on the path from $r_i$ to $a$ and count $\text{value}[u]$ one by one. There are two possible cases:


$\text{value}[u]=\text{value}[u']$ for some $u'\neq u$ on the path from $r_i$ to $u$. In this case, we have already counted $\text{value}[u]$ before so just ignore it.
If not, we should count the number of $b$ such that $\text{value}[u]$ doesn't appear in $\text{path}(r,b)$ and add it to $AR[a]$. This is exactly $(n-n_i)-(|S(\text{value}[u])|-|S_i(\text{value}[u])|)$ where $n_i$ is the number of nodes in the $i$-th branch.


Note that the number we count for each $u$ is independent from $a$. Therefore we can do this for every $a$ in a single DFS which takes $O(n)$. (Tell me if you need more details.)

Now the only remaining part is the preprocessing which computes $|S(v)|$, and it's actually similar to previous steps:

Consider each node $u$. 


If $\text{value}[u]=\text{value}[u']$ for some $u'\neq u$ on the path from $r$ to $u$, we have already counted $\text{value}[u]$ before so just ignore it.
If not, add $n_u$ to $|S(\text{value}[u])|$. Here $n_u$ means the size of subtree rooted at $u$.


$|S_i(v)|$ can be computed similarly. The preprocessing also takes $O(n)$.

The complete flow seems like 

Solve(T):
  Choose a centroid r as the root
  Preprocess() //Compute |S(v)| and |S_i(v)| via a tree traversal

  AR[r]+=Sum{|S(v)|}
  for each node u
    AR[u]+=Sum{|S(v)|-|S_i(v)|}

  Update AR with the second term via another tree traversal

  for each branch T_i of r (which is also a tree)
    Solve(T_i)