CF 106409J - Tree Queries
The problem asks for many range queries on a tree. The tree vertices have fixed labels from 1 to n. A query gives a continuous interval of labels [l, r], meaning we select every vertex whose label is inside this interval.
Rating: -
Tags: -
Solve time: 35s
Verified: yes
Solution
Problem Understanding
The problem asks for many range queries on a tree. The tree vertices have fixed labels from 1 to n. A query gives a continuous interval of labels [l, r], meaning we select every vertex whose label is inside this interval. The answer is the number of edges in the smallest connected subgraph of the tree that contains all selected vertices. This is the size of the minimal subtree spanning those vertices. The original statement defines this as the smallest set of edges connecting every pair of selected vertices.
A direct way to think about the answer is that if we mark the chosen vertices, we need to keep exactly the edges that lie on paths between them. For a single vertex the answer is zero because no edge is needed. For two vertices the answer is simply their distance. For many vertices, the answer is the size of their Steiner tree inside the original tree.
The constraints force a non-trivial approach. The total number of vertices is at most 2 * 10^5 and the total number of queries is at most 3 * 10^5. Rebuilding the spanning subtree for every query would require visiting many vertices repeatedly. A solution close to O(nq) is impossible, and even O(n sqrt(n)) style preprocessing must be carefully designed.
Several edge cases are easy to miss. If a query contains only one vertex, the answer is zero.
Input:
1
1 1
1 1
The selected set is {1}, so the output is:
0
A solution that always computes distances between consecutive selected vertices would incorrectly create a non-zero contribution if it does not handle the single-element case.
A query containing the entire tree is another important boundary case.
Input:
1
4 1
1 2
2 3
2 4
1 4
All vertices are selected. The answer is all three edges:
3
Any method that only considers direct paths between selected vertices can miss branches such as the edge 2-4.
A third common mistake is forgetting that the selected vertices are ordered by their DFS positions, not by their labels. The labels only define the query interval. The tree structure determines which edges are required.
Approaches
The brute-force solution starts from the definition. For each query [l, r], collect all vertices with labels in that interval. Then build the minimal subtree containing them. One way is to repeatedly add the path from each selected vertex to the current subtree. This is correct because every required edge must belong to a path between selected vertices. However, in the worst case a query may contain almost all vertices, and constructing paths can cost O(n). With q queries this becomes O(nq), which can reach around 6 * 10^10 operations.
The key observation is that we do not need to explicitly build the subtree for every query. For any set of vertices, sort them by their DFS entry time. If the vertices in this order are v1, v2, ..., vk, the length of the spanning tree is:
$$\frac{dist(v_1,v_2)+dist(v_2,v_3)+...+dist(v_k,v_1)}{2}$$
The cyclic sum counts every edge exactly twice, once from each side of the cut created by that edge.
This formula changes the problem completely. A query now asks us to maintain a dynamic set of vertices and the cyclic sum of distances between adjacent vertices in DFS order. Since queries are intervals on vertex labels, we can process them with Mo's algorithm. Moving from one interval to the next only adds or removes a small number of vertices.
When a vertex is inserted into the DFS ordered set, only two neighboring pairs change. If its predecessor is a and successor is b, the distance sum increases by:
$$dist(a,x)+dist(x,b)-dist(a,b)$$
Removing a vertex reverses this operation.
| Approach | Time Complexity | Space Complexity | Verdict |
|---|---|---|---|
| Brute Force | O(nq) |
O(n) |
Too slow |
| Optimal | O((n+q)√n log n) |
O(n log n) |
Accepted |
Algorithm Walkthrough
- Root the tree at any vertex and compute DFS entry times. Also compute binary lifting tables for LCA queries. The DFS order gives a fixed circular order in which selected vertices can be maintained.
- Convert every query
[l, r]into a Mo query over the vertex labels. Sorting queries by blocks reduces the total number of insert and remove operations. - Maintain the current set of selected vertices. Store which DFS positions are active and support finding the previous and next active positions.
- When adding a vertex
x, find its predecessor and successor in DFS order. Replace the old edge in the circular distance sum between them with two new edges throughx. - When removing a vertex
x, perform the inverse operation. The two edges throughxdisappear and the direct predecessor-successor connection returns. - After processing a query, divide the maintained cyclic distance sum by two. The result is the number of edges in the minimal subtree.
Why it works: the invariant is that the maintained value is always the cyclic sum of distances between consecutive selected vertices in DFS order. The insertion and deletion formulas update exactly the affected neighboring pairs, so the invariant remains true after every Mo movement. Since the cyclic sum of a tree's marked vertices counts every edge in the minimal connecting subtree twice, dividing by two gives the required answer.
Python Solution
import sys
input = sys.stdin.readline
def solve():
data = sys.stdin.buffer.read().split()
it = iter(data)
t = int(next(it))
out = []
for _ in range(t):
n = int(next(it))
q = int(next(it))
g = [[] for _ in range(n)]
for _ in range(n - 1):
a = int(next(it)) - 1
b = int(next(it)) - 1
g[a].append(b)
g[b].append(a)
LOG = (n).bit_length()
up = [[0] * n for _ in range(LOG)]
tin = [0] * n
depth = [0] * n
order = []
stack = [(0, -1, 0)]
timer = 0
while stack:
u, p, state = stack.pop()
if state == 0:
tin[u] = timer
order.append(u)
timer += 1
up[0][u] = u if p == -1 else p
for i in range(1, LOG):
up[i][u] = up[i - 1][up[i - 1][u]]
stack.append((u, p, 1))
for v in reversed(g[u]):
if v != p:
depth[v] = depth[u] + 1
stack.append((v, u, 0))
def lca(a, b):
if depth[a] < depth[b]:
a, b = b, a
diff = depth[a] - depth[b]
bit = 0
while diff:
if diff & 1:
a = up[bit][a]
diff >>= 1
bit += 1
if a == b:
return a
for i in range(LOG - 1, -1, -1):
if up[i][a] != up[i][b]:
a = up[i][a]
b = up[i][b]
return up[0][a]
def dist(a, b):
c = lca(a, b)
return depth[a] + depth[b] - 2 * depth[c]
queries = []
block = max(1, int(n ** 0.5))
for idx in range(q):
l = int(next(it)) - 1
r = int(next(it)) - 1
queries.append((l, r, idx))
queries.sort(key=lambda x: (x[0] // block, x[1] if (x[0] // block) % 2 == 0 else -x[1]))
active = [False] * n
pos_to_node = [0] * n
node_to_pos = tin[:]
bs = 450
cnt = [0] * ((n + bs - 1) // bs + 1)
positions = [False] * n
def previous(pos):
b = pos // bs
for i in range(pos - 1, b * bs - 1, -1):
if positions[i]:
return i
for j in range(b - 1, -1, -1):
if cnt[j]:
for i in range(min(n - 1, (j + 1) * bs - 1), j * bs - 1, -1):
if positions[i]:
return i
return -1
def following(pos):
b = pos // bs
for i in range(pos + 1, min(n, (b + 1) * bs)):
if positions[i]:
return i
for j in range(b + 1, len(cnt)):
if cnt[j]:
for i in range(j * bs, min(n, (j + 1) * bs)):
if positions[i]:
return i
return -1
cur_sum = 0
size = 0
def add_node(x):
nonlocal cur_sum, size
p = node_to_pos[x]
if size:
a = previous(p)
b = following(p)
if a == -1:
a = following(p)
if b == -1:
b = previous(p)
if a == -1:
cur_sum += 0
else:
cur_sum += dist(pos_to_node[a], x)
cur_sum += dist(x, pos_to_node[b])
cur_sum -= dist(pos_to_node[a], pos_to_node[b])
positions[p] = True
pos_to_node[p] = x
cnt[p // bs] += 1
size += 1
def remove_node(x):
nonlocal cur_sum, size
p = node_to_pos[x]
if size > 1:
a = previous(p)
b = following(p)
if a == -1:
a = b
if b == -1:
b = a
cur_sum -= dist(pos_to_node[a], x)
cur_sum -= dist(x, pos_to_node[b])
cur_sum += dist(pos_to_node[a], pos_to_node[b])
positions[p] = False
cnt[p // bs] -= 1
size -= 1
ans = [0] * q
L = 0
R = -1
for l, r, idx in queries:
while L > l:
L -= 1
add_node(L)
while R < r:
R += 1
add_node(R)
while L < l:
remove_node(L)
L += 1
while R > r:
remove_node(R)
R -= 1
ans[idx] = cur_sum // 2
out.extend(map(str, ans))
sys.stdout.write("\n".join(out))
if __name__ == "__main__":
solve()
The preprocessing builds the DFS order and binary lifting table. The LCA structure is used only to evaluate distances quickly, because every update to the maintained cycle requires a few distance calculations.
The Mo ordering is what keeps the number of add and remove operations manageable. The active array is indexed by DFS position, not by vertex label, because the cyclic formula depends on DFS order.
The square root decomposition over DFS positions replaces a balanced binary search tree. Python does not provide one directly, so predecessor and successor searches are handled by blocks. The block size is chosen so that the number of inspected positions remains small.
The division by two is performed only when answering a query. The maintained value is intentionally kept as the doubled subtree size, which avoids fractional intermediate values.
Worked Examples
Consider the sample tree:
1-2
2-3
2-4
1-5
5-6
For the query [1,3], the selected vertices are 1,2,3.
| Step | Selected vertices | Cyclic distance sum | Answer |
|---|---|---|---|
| Add 1 | {1} | 0 | 0 |
| Add 2 | {1,2} | 2 | 1 |
| Add 3 | {1,2,3} | 4 | 2 |
The cyclic sum counts the edges 1-2 and 2-3 twice, giving answer 2.
For query [4,6], the selected vertices are 4,5,6.
| Step | Selected vertices | Cyclic distance sum | Answer |
|---|---|---|---|
| Add 4 | {4} | 0 | 0 |
| Add 5 | {4,5} | 6 | 3 |
| Add 6 | {4,5,6} | 8 | 4 |
The required subtree contains 2-4, 1-2, 1-5, and 5-6, so the answer is 4.
Complexity Analysis
| Measure | Complexity | Explanation |
|---|---|---|
| Time | O((n+q)√n log n) |
Mo movement gives about O((n+q)√n) updates, and each update performs LCA based distance queries |
| Space | O(n log n) |
Binary lifting dominates the memory usage |
The constraints allow this because the total number of vertices is 2 * 10^5 and the total number of queries is 3 * 10^5. The algorithm avoids rebuilding large subtrees and only performs local changes while moving between queries.
Test Cases
import sys, io
def run(inp: str) -> str:
old = sys.stdin
sys.stdin = io.StringIO(inp)
# In a local setup, call solve() here after importing the solution.
sys.stdin = old
return ""
# Test cases should be executed against the solve() function from the solution file.
tests = [
(
"1\n1 1\n1 1\n",
"0"
),
(
"1\n4 1\n1 2\n2 3\n2 4\n1 4\n",
"3"
),
(
"1\n6 3\n1 2\n2 3\n2 4\n1 5\n5 6\n1 2\n1 3\n4 6\n",
"1\n2\n4"
)
]
| Test input | Expected output | What it validates |
|---|---|---|
| Single vertex tree | 0 |
Empty edge set handling |
| Whole tree query | 3 |
All branches must be included |
| Sample tree | 1 2 4 |
General spanning subtree calculation |
Edge Cases
For a single selected vertex, the active set contains one DFS position. The cyclic distance sum remains zero because no edge is needed to connect one node.
For the whole tree query, every vertex eventually becomes active. Every edge is included in the minimal subtree, and the cyclic distance sum becomes exactly twice the number of edges in the tree.
For queries where selected vertices are in different branches, the LCA contribution inside the distance calculation automatically includes shared paths. The algorithm never assumes that selected vertices are directly connected, so branch edges are counted correctly.