Identity

For distinct vertices $u,v$, define

$$\delta(u,v)=\max(1,d_T(u,v)).$$

This function satisfies the triangle inequality.

Suppose the vertices occur from left to right in the order $p_1,p_2,\ldots,p_N$. Every consecutive coordinate gap is at least . Conversely, setting the consecutive gaps equal to these values satisfies every pairwise constraint by the triangle inequality. The problem is therefore the shortest Hamiltonian path problem in the metric .

Contract all edges of weight $0$. The resulting graph $H$ is a tree whose edges all have weight $1$. Let $c_v$ be the number of original vertices represented by a vertex $v$ of , and let be its degree in .

Expand every transition between different components into its path in $H$. This gives a walk covering $H$. If its first and last components are $s$ and $t$, let $P$ be their path. Every edge on $P$ is traversed at least once, and every other edge at least twice.

Two extra traversals of an edge cost $2$ and add at most one usable visit at each endpoint. Each additional original vertex placed during the same visit also costs $1$, so extra edge traversals cannot improve the answer. It is sufficient to traverse every edge on $P$ once and every other edge twice.

In such a walk, a component $v$ is visited $\deg(v)-1$ times if it is an internal vertex of $P$, and $\deg(v)$ times otherwise. Original vertices exceeding this number cost one each because they must be placed consecutively within their component.

Define

$$B=2(|V(H)|-1)+\sum_{v\in V(H)}\max(0,c_v-\deg(v)).$$

Making $v$ an internal vertex of $P$ adds one to the cost exactly when $c_v\geq\deg(v)$. Call such a vertex marked. The answer for a path $P$ is

$$B-\left(|E(P)|-\#\{\text{marked internal vertices of }P\}\right).$$

Thus, maximize the parenthesized benefit.

Root $H$ arbitrarily. Let $down_v$ be the maximum benefit of a path having $v$ as one endpoint and going into the subtree of $v$. An arm through a child $$ has value

$$1+\max(0,down_u-[u\text{ is marked}]).$$

Keeping the two largest child arms checks both paths ending at $v$ and paths passing through $v$. Process an iterative DFS order in reverse to avoid recursion depth issues.

If $H$ has only one vertex, all pairwise distances are $1$, so the answer is $N-1$.

The time complexity is $O(N\alpha(N))$, and the space complexity is $O(N)$.

In subtask 1, if the tree diameter is $D$, the answer is $2(N-1)-D$. In subtask 2, enumerate the starting vertex and traverse the contracted tree to evaluate every path benefit in $O(N^2)$ time.