#P1044F. DFS
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ID: 4408
Type: RemoteJudge
6000ms
512MiB
Tried: 0
Accepted: 0
Difficulty: (None)
Uploaded By:
Hydro
Tags> DFS
No submission language available for this problem.
Description
Let $T$ be a tree on $n$ vertices. Consider a graph $G_0$, initially equal to $T$. You are given a sequence of $q$ updates, where the $i$-th update is given as a pair of two distinct integers $u_i$ and $v_i$.
For every $i$ from $1$ to $q$, we define the graph $G_i$ as follows:
- If $G_{i-1}$ contains an edge $\{u_i, v_i\}$, then remove this edge to form $G_i$.
- Otherwise, add this edge to $G_{i-1}$ to form $G_i$.
Formally, $G_i := G_{i-1} \triangle \{\{u_i, v_i\}\}$ where $\triangle$ denotes the set symmetric difference.
Furthermore, it is guaranteed that $T$ is always a subgraph of $G_i$. In other words, an update never removes an edge of $T$.
Consider a connected graph $H$ and run a depth-first search on it. One can see that the tree edges (i.e. the edges leading to a not yet visited vertex at the time of traversal) form a spanning tree of the graph $H$. This spanning tree is not generally fixed for a particular graph — it depends on the starting vertex, and on the order in which the neighbors of each vertex are traversed.
We call vertex $w$ good if one can order the neighbors of each vertex in such a way that the depth-first search started from $w$ produces $T$ as the spanning tree. For every $i$ from $1$ to $q$, find and report the number of good vertices.
The first line contains two integers $n$ and $q$ ($3 \le n \le 2\cdot 10^5$, $1 \le q \le 2 \cdot 10^5$) — the number of nodes and the number of updates, respectively.
Each of the next $n-1$ lines contains two integers $u$ and $v$ ($1 \le u, v \le n$, $u \ne v$) — vertices connected by an edge in $T$. It is guaranteed that this graph is a tree.
Each of the next $q$ lines contains two integers $u$ and $v$ ($1 \le u, v \le n$, $u \ne v$) — the endpoints of the edge that is added or removed. It is guaranteed that this edge does not belong to $T$.
For each update, print one integer $k$ — the number of good vertices $w$ after the corresponding update.
Input
The first line contains two integers $n$ and $q$ ($3 \le n \le 2\cdot 10^5$, $1 \le q \le 2 \cdot 10^5$) — the number of nodes and the number of updates, respectively.
Each of the next $n-1$ lines contains two integers $u$ and $v$ ($1 \le u, v \le n$, $u \ne v$) — vertices connected by an edge in $T$. It is guaranteed that this graph is a tree.
Each of the next $q$ lines contains two integers $u$ and $v$ ($1 \le u, v \le n$, $u \ne v$) — the endpoints of the edge that is added or removed. It is guaranteed that this edge does not belong to $T$.
Output
For each update, print one integer $k$ — the number of good vertices $w$ after the corresponding update.
Samples
3 2
1 2
1 3
2 3
3 2
2
3
6 6
1 2
2 3
1 4
4 5
1 6
2 5
3 4
5 2
6 4
3 4
6 5
3
2
3
2
3
2
Note
The first sample is depicted in the following figure.
After the first update, $G$ contains all three possible edges. The result of a DFS is as follows:
- Let the starting vertex be $1$. We have two choices of ordering the neighbors of $1$, either $[2, 3]$ or $[3, 2]$.
- If we choose the former, then we reach vertex $2$. Regardless of the ordering of its neighbors, the next visited vertex will be $3$. Thus, the spanning tree generated by this DFS will contain edges $\{1, 2\}$ and $\{2, 3\}$, which does not equal to $T$.
- If we choose the latter, we obtain a spanning tree with edges $\{1, 3\}$ and $\{2, 3\}$.
- Let the starting vertex be $2$. We have two choices of traversing its neighbors. If we visit $3$ first, then the spanning tree will consist of edges $\{2,3\}$ and $\{1,3\}$, which is not equal to $T$. If we, however, visit $1$ first, then we can only continue to $3$ from here, and the spanning tree will consist of edges $\{1, 2\}$ and $\{1,3\}$, which equals to $T$. Hence, $2$ is a good vertex.
- The case when we start in the vertex $3$ is symmetrical to starting in $2$, and hence $3$ is a good vertex.
After the second update, the edge between $2$ and $3$ is removed, and $G = T$. It follows that the spanning tree generated by DFS will be always equal to $T$ independent of the choice of the starting vertex. Thus, the answer is $3$.
In the second sample, the set of good vertices after the corresponding query is:
- $\{2, 3, 5\}$
- $\{3, 5\}$
- $\{3, 4, 5\}$
- $\{4, 5\}$
- $\{4, 5, 6\}$
- $\{5, 6\}$