CF 2218F - The 67th Tree Problem
We are asked to build a rooted tree with exactly $n = x + y$ vertices. The root is fixed to be vertex $1$. For every vertex, consider the size of its rooted subtree.
CF 2218F - The 67th Tree Problem
Rating: 1500
Tags: constructive algorithms, implementation, trees
Solve time: 3m 38s
Verified: no
Solution
Problem Understanding
We are asked to build a rooted tree with exactly $n = x + y$ vertices. The root is fixed to be vertex $1$.
For every vertex, consider the size of its rooted subtree. A vertex is counted in the "even" group if its subtree contains an even number of vertices, and in the "odd" group otherwise.
The goal is to construct any tree with exactly $x$ even-subtree vertices and exactly $y$ odd-subtree vertices, or determine that no such tree exists.
The total number of vertices over all test cases is at most $2 \cdot 10^5$. That immediately tells us that any accepted solution must run in linear time per test case, or linear in the total input size. Anything that tries to search over tree shapes or perform expensive combinatorial generation is completely infeasible.
The subtle part of the problem is not building a tree. The real challenge is understanding which pairs $(x,y)$ are even possible.
Consider a few small examples.
For $n=1$, the only tree consists of the root. Its subtree size is $1$, so the counts are $(x,y)=(0,1)$.
For $n=2$, the root has subtree size $2$ and the leaf has subtree size $1$. The counts are $(1,1)$.
For $n=3$, a star gives $(0,3)$, while a chain gives $(1,2)$. No tree gives $(2,1)$.
A naive attempt might assume that any split of $n$ into even and odd counts is achievable. The example $(x,y)=(2,1)$ already disproves that.
Another easy mistake is forgetting that the root's subtree size is always $n$. If $n$ is even, the root is automatically an even-subtree vertex. For example:
x = 0, y = 4
would require every vertex to have odd subtree size, but the root must be even because the whole tree contains $4$ vertices. The correct answer is NO.
The key task is to find the exact feasibility condition and then construct a tree that meets it.
Approaches
A brute-force approach would try to enumerate tree structures on $n$ vertices, compute every subtree size, and check whether the resulting counts match $(x,y)$.
This works conceptually because subtree sizes completely determine whether a tree is valid. The problem is that the number of trees grows exponentially. Even for a few dozen vertices, enumeration becomes impossible. With $n$ up to $2 \cdot 10^5$, brute force is not remotely viable.
The breakthrough comes from studying parity instead of exact subtree sizes.
Let an odd node mean a vertex whose subtree size is odd, and an even node mean a vertex whose subtree size is even.
For any vertex with even subtree size, the parity equation
$$\text{subtree}(v) \equiv 1 + \sum \text{subtree}(child) \pmod 2$$
implies that the number of odd children must be odd. In particular, every even node has at least one odd child.
This observation creates an injection from even vertices to odd vertices. Assign each even vertex one odd child. Since every odd vertex has only one parent, the same odd vertex cannot be assigned to two different even vertices.
When $n$ is odd, the root is odd. The injection goes into the non-root odd vertices, giving
$$x \le y - 1.$$
Since $n=x+y$,
$$2x \le n-1.$$
When $n$ is even, the root is even, so the injection goes into all odd vertices:
$$x \le y.$$
Since $n=x+y$,
$$2x \le n.$$
Both cases simplify to
$$x \le \left\lfloor \frac n2 \right\rfloor.$$
There is one extra restriction. If $n$ is even, the root itself is even, so $x=0$ is impossible.
The remarkable part is that these conditions are also sufficient. Once we know that, the whole problem becomes a constructive exercise.
| Approach | Time Complexity | Space Complexity | Verdict |
|---|---|---|---|
| Brute Force | Exponential | Exponential | Too slow |
| Optimal | O(n) | O(n) | Accepted |
Algorithm Walkthrough
Feasibility Check
Let
$$n = x + y.$$
A solution exists if and only if:
$$x \le \left\lfloor \frac n2 \right\rfloor$$
and additionally
$$(n \text{ even}) \implies (x \ge 1).$$
If either condition fails, output NO.
Construction for Odd n
When $n$ is odd:
- If $x=0$, output a star centered at vertex $1$.
- Otherwise create $x$ disjoint paths of length $2$:
$$1 - a_i - b_i.$$ 3. Attach every remaining vertex directly to the root.
Each middle vertex $a_i$ has subtree size $2$, so it is even.
The root is odd because $n$ is odd.
No other vertex is even.
Hence the number of even vertices is exactly $x$.
Construction for Even n
When $n$ is even:
- The root is automatically even.
- We still need $x-1$ additional even vertices.
- Create $x-1$ disjoint paths of length $2$:
$$1 - a_i - b_i.$$ 4. Attach every remaining vertex directly to the root.
Each $a_i$ contributes one even vertex.
Together with the root, the total number of even vertices becomes exactly $x$.
Why it works
Every path $1-a_i-b_i$ contributes exactly one non-root even vertex, namely $a_i$, because its subtree contains exactly two vertices.
Every leaf has subtree size $1$, so it is odd.
All extra vertices attached directly to the root are leaves and remain odd.
For odd $n$, the root has odd subtree size and contributes no even count. The $x$ middle vertices are the only even vertices.
For even $n$, the root contributes one even vertex and the $x-1$ middle vertices contribute the remaining even vertices.
The construction never creates any unexpected even nodes, so the count is exact.
Python Solution
import sys
input = sys.stdin.readline
t = int(input())
out = []
for _ in range(t):
x, y = map(int, input().split())
n = x + y
if x > n // 2:
out.append("NO")
continue
if n % 2 == 0 and x == 0:
out.append("NO")
continue
edges = []
nxt = 2
if n % 2 == 1:
pairs = x
else:
pairs = x - 1
for _ in range(pairs):
a = nxt
b = nxt + 1
nxt += 2
edges.append((1, a))
edges.append((a, b))
while nxt <= n:
edges.append((1, nxt))
nxt += 1
out.append("YES")
for u, v in edges:
out.append(f"{u} {v}")
sys.stdout.write("\n".join(out))
The first part checks the feasibility condition derived from the parity argument.
The variable pairs stores the number of length-two branches that must be created. For odd n, every such branch contributes one even vertex, so we need exactly x branches. For even n, the root already contributes one even vertex, so only x - 1 branches are needed.
Each branch consumes two new vertices. After creating all required branches, every remaining vertex is attached directly to the root as a leaf.
The numbering is handled sequentially with nxt, which guarantees that exactly n vertices are used and exactly n-1 edges are produced.
A common off-by-one mistake is forgetting that the root already counts as an even vertex when n is even. Using pairs = x in both cases would produce one extra even vertex.
Worked Examples
Example 1
Input:
x = 1, y = 2
Then:
$$n = 3$$
which is odd.
| Step | Value |
|---|---|
| n | 3 |
| pairs | 1 |
| Create branch | 1-2-3 |
| Remaining vertices | none |
Constructed tree:
1
|
2
|
3
Subtree sizes:
| Vertex | Subtree Size | Parity |
|---|---|---|
| 1 | 3 | Odd |
| 2 | 2 | Even |
| 3 | 1 | Odd |
We obtain exactly one even vertex and two odd vertices.
Example 2
Input:
x = 4, y = 7
Then:
$$n = 11$$
which is odd.
| Step | Value |
|---|---|
| n | 11 |
| pairs | 4 |
| Used vertices | 2..9 |
| Remaining vertices | 10, 11 |
Generated edges:
1-2-3
1-4-5
1-6-7
1-8-9
1-10
1-11
Parity table:
| Vertex Type | Count |
|---|---|
| Middle vertices of branches | 4 even |
| Root | 1 odd |
| Leaves | 6 odd |
Total:
| Even | Odd |
|---|---|
| 4 | 7 |
The target counts are matched exactly.
Complexity Analysis
| Measure | Complexity | Explanation |
|---|---|---|
| Time | O(n) | Each vertex and edge is generated once |
| Space | O(n) | The edge list contains exactly n−1 edges |
Since the sum of all $n=x+y$ across test cases is at most $2\cdot10^5$, the total work is linear in the input size. This easily fits within both the 4-second time limit and the memory limit.
Test Cases
# helper validator for produced constructions
import sys
import io
from collections import deque
def validate(x, y, edges):
n = x + y
if n == 1:
return x == 0 and y == 1
g = [[] for _ in range(n + 1)]
for u, v in edges:
g[u].append(v)
g[v].append(u)
parent = [0] * (n + 1)
order = [1]
for v in order:
for to in g[v]:
if to != parent[v]:
parent[to] = v
order.append(to)
if len(order) != n:
return False
sz = [1] * (n + 1)
for v in reversed(order):
if parent[v]:
sz[parent[v]] += sz[v]
even = sum(s % 2 == 0 for s in sz[1:])
odd = n - even
return even == x and odd == y
# feasibility checks
assert (0 <= (1 // 2)) and (1 % 2 == 1)
assert not (0 > (1 // 2))
# minimum size
assert (0, 1) == (0, 1)
# impossible: root must be even
assert (0 > 4 // 2) is False
# boundary equality
assert 3 == 6 // 2
# maximum-style boundary
n = 200000
assert n // 2 == 100000
The exact output tree is not unique, so assertion-based testing is best done with a validator that checks subtree parities rather than comparing against one fixed output.
| Test input | Expected output | What it validates |
|---|---|---|
| x=0, y=1 | YES | Smallest possible tree |
| x=0, y=4 | NO | Even-sized tree cannot have zero even vertices |
| x=3, y=3 | YES | Boundary case x=n/2 |
| x=100000, y=100000 | YES | Maximum total size |
| x=2, y=1 | NO | Exceeds theoretical maximum number of even vertices |
Edge Cases
Consider:
x = 0, y = 4
Here $n=4$ is even. The root's subtree contains all four vertices, so the root is even. Having zero even vertices is impossible. The algorithm detects n % 2 == 0 and x == 0 and immediately outputs NO.
Consider:
x = 2, y = 1
Here $n=3$. The feasibility bound requires
$$x \le \left\lfloor \frac 32 \right\rfloor = 1.$$
Since $x=2$, the request exceeds the maximum achievable number of even-subtree vertices. The algorithm outputs NO.
Consider:
x = 2, y = 3
Here $n=5$ and $x=n//2$, the largest possible valid value. The construction creates two length-two branches and no extra leaves:
1-2-3
1-4-5
Vertices $2$ and $4$ have subtree size $2$, giving exactly two even vertices. This shows that the upper bound is tight.
Consider:
x = 0, y = 5
Here $n=5$ is odd. The algorithm outputs a star. Every leaf has subtree size $1$, and the root has subtree size $5$. All subtree sizes are odd, so the counts are exactly $(0,5)$. This is the unique situation where zero even vertices is possible.