You are given a tree (an acyclic undirected connected graph) with N nodes. The tree nodes are numbered from 1 to N. In the start, the color of any node in the tree is white.
We will ask you to perfrom some instructions of the following form:
- 0 i : change the color of the i-th node (from white to black, or from black to white);
or - 1 v : ask for the id of the first black node on the path from node 1 to node v. if it doesn‘t exist, you may return -1 as its result.
Input
In the first line there are two integers N and Q.
In the next N-1 lines describe the edges in the tree: a line with two integers a b denotes an edge between a and b.
The next Q lines contain instructions "0 i" or "1 v" (1 ≤ i, v ≤ N).
Output
For each "1 v" operation, write one integer representing its result.
Example
Input: 9 8 1 2 1 3 2 4 2 9 5 9 7 9 8 9 6 8 1 3 0 8 1 6 1 7 0 2 1 9 0 2 1 9 Output: -1 8 -1 2 -1
Constraints & Limits
There are 12 real input files.
For 1/3 of the test cases, N=5000, Q=400000.
For 1/3 of the test cases, N=10000, Q=300000.
For 1/3 of the test cases, N=100000, Q=100000.
一棵树，点数<=100000，初始全是白点，支持两种操作：
1.将某个点的颜色反色。
2.询问某个点至根节点路径上第一个黑点是哪个。
树链剖分
注意到询问的链都是(1,v)
在线段树上维护区间内深度最浅的黑色点位置即可，注意线段树结点到原树的反向映射
1 #include<iostream>
2 #include<cstdio>
3 #include<algorithm>
4 #include<cstring>
5 #include<queue>
6 using namespace std;
7 const int INF=1e9;
8 const int mxn=100010;
9 int read(){
10 int x=0,f=1;char ch=getchar();
11 while(ch<‘0‘ || ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
12 while(ch>=‘0‘ && ch<=‘9‘){x=x*10-‘0‘+ch;ch=getchar();}
13 return x*f;
14 }
15 struct edge{int v,nxt;}e[mxn<<1];
16 int hd[mxn],mct=0;
17 int n,q;
18 void add_edge(int u,int v){
19 e[++mct].v=v;e[mct].nxt=hd[u];hd[u]=mct;return;
20 }
21 struct node{
22 int fa,son;
23 int top,size;
24 int w;
25 }t[mxn];
26 int sz=0;
27 int dep[mxn];
28 int id[mxn];
29 void DFS1(int u,int fa){
30 dep[u]=dep[fa]+1;
31 t[u].size=1;
32 for(int i=hd[u];i;i=e[i].nxt){
33 int v=e[i].v;if(v==fa)continue;
34 t[v].fa=u;
35 DFS1(v,u);
36 t[u].size+=t[v].size;
37 if(t[v].size>t[t[u].son].size)
38 t[u].son=v;
39 }
40 return;
41 }
42 void DFS2(int u,int top){
43 t[u].w=++sz;t[u].top=top;
44 id[sz]=u;
45 if(t[u].son)DFS2(t[u].son,top);
46 for(int i=hd[u];i;i=e[i].nxt){
47 int v=e[i].v;
48 if(v==t[u].fa || v==t[u].son)continue;
49 DFS2(v,v);
50 }
51 return;
52 }
53 struct SGT{
54 int ps;
55 int c;
56 }st[mxn<<2];
57 void Build(int l,int r,int rt){
58 st[rt].ps=INF;
59 if(l==r)return;
60 int mid=(l+r)>>1;
61 Build(l,mid,rt<<1);Build(mid+1,r,rt<<1|1);
62 return;
63 }
64 void update(int p,int l,int r,int rt){
65 if(l==r){
66 st[rt].c^=1;
67 st[rt].ps=(st[rt].c)?l:INF;
68 return;
69 }
70 int mid=(l+r)>>1;
71 if(p<=mid)update(p,l,mid,rt<<1);
72 else update(p,mid+1,r,rt<<1|1);
73 st[rt].ps=min(st[rt<<1].ps,st[rt<<1|1].ps);
74 return;
75 }
76 int query(int L,int R,int l,int r,int rt){
77 if(L<=l && r<=R){return st[rt].ps;}
78 int mid=(l+r)>>1;
79 int res=INF;
80 if(L<=mid)res=min(res,query(L,R,l,mid,rt<<1));
81 if(R>mid && res>mid)res=min(res,query(L,R,mid+1,r,rt<<1|1));
82 return res;
83 }
84 int Qt(int x,int y){
85 int res=INF;
86 while(t[x].top!=t[y].top){
87 if(dep[t[x].top]<dep[t[y].top])swap(x,y);
88 res=min(res,query(t[t[x].top].w,t[x].w,1,n,1));
89 x=t[t[x].top].fa;
90 }
91 if(dep[x]>dep[y])swap(x,y);
92 res=min(res,query(t[x].w,t[y].w,1,n,1));
93 return res;
94 }
95 int main(){
96 int i,j,u,v;
97 n=read();q=read();
98 for(i=1;i<n;i++){
99 u=read();v=read();
100 add_edge(u,v);
101 add_edge(v,u);
102 }
103 DFS1(1,0);
104 DFS2(1,1);
105 Build(1,n,1);
106 while(q--){
107 u=read();v=read();
108 if(!u)
109 update(t[v].w,1,n,1);
110 else{
111 int ans=Qt(1,v);
112 if(ans>=INF){printf("-1\n");continue;}
113 ans=id[ans];
114 printf("%d\n",ans);
115 }
116 }
117 return 0;
118 }