SPOJ QTREE3 - Query on a tree again!

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

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

 #include<iostream>

 #include<cstdio>

 #include<algorithm>

 #include<cstring>

 #include<queue>

 using namespace std;

 const int INF=1e9;

 const int mxn=;

 int read(){

     int x=,f=;char ch=getchar();

     while(ch<'' || ch>''){if(ch=='-')f=-;ch=getchar();}

     while(ch>='' && ch<=''){x=x*-''+ch;ch=getchar();}

     return x*f;

 }

 struct edge{int v,nxt;}e[mxn<<];

 int hd[mxn],mct=;

 int n,q;

 void add_edge(int u,int v){

     e[++mct].v=v;e[mct].nxt=hd[u];hd[u]=mct;return;

 }

 struct node{

     int fa,son;

     int top,size;

     int w;

 }t[mxn];

 int sz=;

 int dep[mxn];

 int id[mxn];

 void DFS1(int u,int fa){

     dep[u]=dep[fa]+;

     t[u].size=;

     for(int i=hd[u];i;i=e[i].nxt){

         int v=e[i].v;if(v==fa)continue;

         t[v].fa=u;

         DFS1(v,u);

         t[u].size+=t[v].size;

         if(t[v].size>t[t[u].son].size)

             t[u].son=v;

     }

     return;

 }

 void DFS2(int u,int top){

     t[u].w=++sz;t[u].top=top;

     id[sz]=u;

     if(t[u].son)DFS2(t[u].son,top);

     for(int i=hd[u];i;i=e[i].nxt){

         int v=e[i].v;

         if(v==t[u].fa || v==t[u].son)continue;

         DFS2(v,v);

     }

     return;

 }

 struct SGT{

     int ps;

     int c;

 }st[mxn<<];

 void Build(int l,int r,int rt){

     st[rt].ps=INF;

     if(l==r)return;

     int mid=(l+r)>>;

     Build(l,mid,rt<<);Build(mid+,r,rt<<|);

     return;

 }

 void update(int p,int l,int r,int rt){

     if(l==r){

         st[rt].c^=;

         st[rt].ps=(st[rt].c)?l:INF;

         return;

     }

     int mid=(l+r)>>;

     if(p<=mid)update(p,l,mid,rt<<);

     else update(p,mid+,r,rt<<|);

     st[rt].ps=min(st[rt<<].ps,st[rt<<|].ps);

     return;

 }

 int query(int L,int R,int l,int r,int rt){

     if(L<=l && r<=R){return st[rt].ps;}

     int mid=(l+r)>>;

     int res=INF;

     if(L<=mid)res=min(res,query(L,R,l,mid,rt<<));

     if(R>mid && res>mid)res=min(res,query(L,R,mid+,r,rt<<|));

     return res;

 }

 int Qt(int x,int y){

     int res=INF;

     while(t[x].top!=t[y].top){

         if(dep[t[x].top]<dep[t[y].top])swap(x,y);

         res=min(res,query(t[t[x].top].w,t[x].w,,n,));

         x=t[t[x].top].fa;

     }

     if(dep[x]>dep[y])swap(x,y);

     res=min(res,query(t[x].w,t[y].w,,n,));

     return res;

 }

 int main(){

     int i,j,u,v;

     n=read();q=read();

     for(i=;i<n;i++){

         u=read();v=read();

         add_edge(u,v);

         add_edge(v,u);

     }

     DFS1(,);

     DFS2(,);

     Build(,n,);

     while(q--){

         u=read();v=read();

         if(!u)

             update(t[v].w,,n,);

         else{

             int ans=Qt(,v);

             if(ans>=INF){printf("-1\n");continue;}

             ans=id[ans];

             printf("%d\n",ans);

         }

     }

     return ;

 }