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 $O(1)$.

You have to find the maximum total weight of two non-intersecting paths.

Once again, choose any two paths and try to "improve" them.

There are two cases to consider: - a path is the diameter and the other path lies completely in a component; - a path is a prefix of the diameter + a vertical path in the corresponding component, the other path is a suffix of the diameter + a vertical path in the corresponding component.

Can you solve them in $O(n)$?

You have to find the maximum total weight of two non-intersecting paths. Notice that the facts stated previously hold even if weights are on nodes instead of edges.

There are two cases to consider: - a path is the diameter and the other path lies completely in a component; - a path is a prefix of the diameter + a vertical path in the corresponding component, the other path is a suffix of the diameter + a vertical path in the corresponding component.

Sketch of proof: suppose that the intersection of a chosen path $i \rightarrow j$ with the diameter is a path $u \rightarrow v$ ($a, u, v, b$ in this order). The paths $a \rightarrow j$, $i \rightarrow b$, $a \rightarrow b$ are longer than $i \rightarrow j$, so it's optimal to take one them instead of $i \rightarrow j$ if possible.

It's possible to solve both cases in $O(n)$: - find the diameter of each component, pick the maximum; - precalculate the height of each component, the distance of each node from $a$, $b$, the prefix maximums $\text{pm}(i)$ of $\text{dist}(a, i) + \text{height}(i)$, the suffix maximums $\text{sm}(i)$ of $\text{dist}(i, b) + \text{height}(i)$, and find the maximum $\text{pm}(i) + \text{sm}(i + 1)$.

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 $dp_{n,n}$.