Hints: Graph Cutting (Two Vertices, Fixed Size)

In the star example ($1$ connected to $2,3,4$) with $a=2$, $b=3$, $d=3$: the path $2 \text{–} 1 \text{–} 3$ already uses $3$ vertices. Can leaf $4$ still be added? What is the answer for this test?

Answer to Hint 1: No — adding $4$ would make size $4 > d$. The only valid subgraph is $\{2,1,3\}$, so the answer is $1$.

Let $L = |P(a,b)|$. Any valid $S$ has $|S| = L + t$ where $t \ge 0$ counts vertices not on the path. Rewrite the problem: count ways to pick exactly $t = d - L$ extra vertices from side branches.

Answer to Hint 3: If $d < L$, the answer is $0$. Otherwise you need the number of ways to choose $t = d - L$ vertices off the path while staying connected (equivalently: each used side branch contributes a connected block containing its attachment point).

For a side branch rooted at $x$, define a polynomial $F_x(z) = \sum_{s \ge 1} \mathrm{ways}_s(x)\, z^s$, where $\mathrm{ways}_s(x)$ counts connected subsets of $x$’s subtree that contain $x$ and have exactly $s$ vertices. What is $F_x(z)$ for a leaf $x$?

Answer to Hint 5: $F_x(z)=z$ (only subset $\{x\}$, size $1$). For an internal side node, each side child $y$ is either skipped ($1$) or contributes $F_y(z)$, so

$$F_x(z)=z\cdot\prod_{y}\bigl(1+F_y(z)\bigr).$$

(The leading $z$ is “include $x$”.)

At a path vertex $v$, side branches are independent. If branches $x_1,\ldots,x_m$ hang off $v$, how do you encode “pick $0$ or more vertices from these branches, staying connected through $v$” as a polynomial in $z$?

Answer to Hint 7: Each branch contributes factor $(1+F_{x_i}(z))$ — the $1$ is “use nothing from this branch”. Multiply:

$$G_v(z)=\prod_{x \text{ side of } v}\bigl(1+F_x(z)\bigr).$$

Coefficient $[z^t]G_v(z)$ counts ways to add exactly $t$ vertices at $v$.

Path vertices $v_0,\ldots,v_{L-1}$ have disjoint side regions. How do you combine $G_{v_0},\ldots,G_{v_{L-1}}$ to count total extras $t$ off the whole path?

Answer to Hint 9: Multiply the polynomials:

$$G(z)=\prod_{i=0}^{L-1} G_{v_i}(z).$$

The answer is $[z^{d-L}]\,G(z)$ (or $0$ if $d-L<0$). Each convolution is over sizes, so complexity is $O(n d^2)$ per test case with $\sum n \le 2000$.

Bridge to E2. E2 asks for pairs $(a,b)$ whose minimal connected subgraph has size $d$ (Fedya’s rule in the original E, but with two vertices). That minimal subgraph is just $P(a,b)$, so $|P(a,b)|=d$. You could run this E1 DP for every pair — what is the slower bound, and what is the direct graph formula?

Answer to Hint 11: Brute force: $O(n^2)$ pairs times $O(n d^2)$ per E1 $\Rightarrow O(n^3 d^2)$. Directly: count unordered pairs with graph distance $d-1$, in $O(n^2)$ BFS or faster with one DFS from each node. No side-branch DP is needed for E2 — but the size-tracking DP here is what generalizes to three vertices in E.