First, put the shops on any path of (unweighted) length $$$k$$$. Can you choose a better path?

The optimal path lies completely on the (weighted) diameter.

Try all possible paths in $$$O(n)$$$.

The optimal path lies completely on the (weighted) diameter. Sketch of proof: suppose that the intersection of the chosen path with the diameter is a path $$$u \rightarrow v$$$ ($$$a, u, v, b$$$ in this order). Then, the maximum distance between a node and the nearest shop is one of the following: $$$\text{dist}(a, u)$$$, $$$\text{dist}(v, b)$$$, $$$\text{dist}(c, d)$$$ ($$$d$$$ on the diameter, $$$d$$$ strictly between $$$u$$$ and $$$v$$$, $$$c$$$ in the component of $$$d$$$). Then, $$$u \rightarrow v$$$ itself is optimal.

Now let's try all paths $$$u \rightarrow v$$$ of maximum (unweighted) length (there are $$$O(n)$$$ of them). Notice that the required distance is also the maximum of $$$\text{dist}(a, u)$$$, $$$\text{dist}(v, b)$$$, $$$\text{dist}(c, d)$$$ ($$$d$$$ on the diameter, but not necessarily between $$$u$$$ and $$$v$$$). If we precalculate the distance of each node from the root of its component and from $$$a$$$, $$$b$$$, we can get the maximum distance between a node and the nearest shop in $$$O(1)$$$.

If there are two adjacent elements, it's optimal to remove them.

In Proof 1, we've seen that a swap changes the number of inversions by $$$1$$$. Here, you should calculate something slightly different than the number of inversions.

Consider two pairs of equal elements. What should their relative positions be at the end?

The answer is $$$n + \text{# of pairs of integers (x, y) such that their order in the array is xyxy}$$$. Can you calculate it efficiently?

Consider two pairs of equal elements (integers $$$(x, y)$$$). If you want to remove those pairs, in some moment their relative positions should be $$$\text{xyyx}$$$ (or $$$\text{yxxy}$$$). Instead, if the relative positions are $$$\text{xyxy}$$$, you want to swap two of these elements. If $$$k$$$ is the number of pairs of integers $$$(x, y)$$$ such that their order in the array is $$$\text{xyxy}$$$, do you need exactly $$$k$$$ swaps (and $$$n$$$ removals)? It turns out that the answer is yes. Sketch of a proof:

• if there are no $$$\text{xyxy}$$$, there are two adjacent equal elements (so you can remove them)
• if you remove two adjacent equal elements, the number of $$$\text{xyxy}$$$ doesn't change
• if there are no adjacent equal elements, there is a $$$\text{xyxy}$$$ such that two of these characters are adjacent (so you can swap them and eliminate a $$$\text{xyxy}$$$)

Now let's calculate $$$k$$$. Let $$$(x_i, y_i)$$$ be $$$n$$$ pairs such that $$$x_i < y_i, a_{x_i} = a_{y_i}$$$: now you have to calculate, for each $$$j$$$, how many indices $$$i$$$ meet $$$x_i < x_j, x_j < y_i < y_j$$$.
You can use a Fenwick tree that contains how many times each value in $$$[1, n]$$$ occurs as a $$$y_i$$$, considering only pairs such that $$$x_i < x_j$$$.
So, for each $$$j$$$, the queries look like

• $$$res := res + \text{range_sum}(x_j + 1, y_j - 1)$$$
• add $$$1$$$ in the position $$$y_j$$$ of the Fenwick tree

When is the answer -1?

The answer is -1 if there are $$$\geq 2$$$ characters with an odd number of occurrences in the string. Otherwise, can you calculate the optimal final position of each character, like in 1430E - String Reversal?

Once again, it's useless to swap equal characters. So, what's the relative position of equal characters? Which characters should be located in symmetrical positions (i.e. $$$i, n - i + 1$$$)?

The $$$i$$$-th leftmost occurrence and the $$$i$$$-th rightmost occurrence of a character $$$c$$$ in the initial string should move to symmetrical positions in the final string. Let's consider such "symmetrical pairs": what's the optimal relative order of two pairs with distinct letters? What about the letter in the center of the string (if its length is odd)?

If there are $$$\geq 2$$$ characters with an odd number of occurrences in the string, you can't make a palindrome.
Since it's useless to swap equal characters, you know which pairs of characters will stay on symmetrical positions in the final string, i.e. the $$$i$$$-th leftmost occurrence and the $$$i$$$-th rightmost occurrence of any character $$$c$$$.
In the string acabcaaababbc, for example, you can make these "symmetrical pairs": positions $$$(1, 10), (2, 13), (3, 8), (4, 12), (6, 7), (9, 11)$$$. The character in position $$$5$$$ should go in the center of the final string (i.e. position $$$7$$$).
The symmetrical pairs have to be nested inside each other, in some order. Given two symmetrical pairs, which of them should be located outside the other one?
It turns out that the pair that contains the leftmost element should be located outside. In fact, if you want to reach $$$\text{xyyx}$$$, the initial configurations with $$$x$$$ on the left ($$$\text{xxyy}$$$, $$$\text{xyxy}$$$, $$$\text{xyyx}$$$) require $$$2, 1, 0$$$ moves respectively, while reaching $$$\text{yxxy}$$$ requires $$$2, 1, 2$$$ moves respectively.
So you can sort the symmetrical pairs by leftmost element and construct the array $$$a$$$ such that $$$a_i$$$ is the final position of $$$s_i$$$ in the palindromic string (in this case, $$$[1, 2, 3, 4, 7, 5, 9, 11, 6, 13, 8, 10, 12]$$$) and the result is the number of inversions of $$$a$$$.

Which balls can you put at the end?

At the end, you have a ball with number $$$n$$$. For example, if it's white, you have to sort the remaining $$$n-1$$$ white balls and $$$n$$$ black balls. How to calculate efficiently the cost required?

Let $$$dp_{i,j}$$$ be the minimum cost required to sort the first $$$i$$$ white balls (with a number from $$$1$$$ to $$$i$$$) and the first $$$j$$$ black balls (with a number from $$$1$$$ to $$$j$$$). You need transitions in $$$O(1)$$$.

Re-read Proof 2 in this tutorial.

It's easy to prove that the leftmost $$$k$$$ balls in the final arrangement are the white balls with a number in $$$[1, i]$$$ and the black balls with a number in $$$[1, j]$$$ for some $$$i, j$$$ such that $$$i+j = k$$$.
For fixed $$$i, j$$$, let $$$dp_{i,j}$$$ be the minimum cost (considering only such balls). If the last ball is white, you can move it immediately to the end (see Proof 2), and the cost is the number of $$$l$$$ in the initial array such that

• $$$l \leq i - 1$$$ if the ball is white, $$$l \leq j$$$ if the ball is black
• $$$\text{pos}(l) > \text{pos}(i)$$$

Then you spend $$$dp_{i-1,j}$$$ to sort the rest of the array. You can calculate the cost similarly if the last ball is black.
Computing such transitions in $$$O(1)$$$ can be done with simple prefix sums (for example, let $$$ps_{i, c, j}$$$ be the number of balls with position $$$\geq i$$$, color $$$c$$$ and value $$$\leq j$$$).
The result is $$$dp_{n,n}$$$.