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.

樹鍊剖分，維護路徑上深度最小的黑點。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int fir[],ne[],to[],
val[],
dep[],fa[],size[],
son[],pos[],top[],
n,q,clo;
void add(int num,int u,int v)
{
    ne[num]=fir[u];
    fir[u]=num;
    to[num]=v;
}
void dfs1(int u)
{
    int v;
    size[u]=;
    for (int i=fir[u];i;i=ne[i])
        if ((v=to[i])!=fa[u])
        {
            dep[v]=dep[u]+;
            fa[v]=u;
            dfs1(v);
            size[u]+=size[v];
            if (!son[u]||size[son[u]]<size[v]) son[u]=v;
        }
}
void dfs2(int u)
{
    int v;
    pos[u]=++clo;
    if (son[u])
    {
        top[son[u]]=top[u];
        dfs2(son[u]);
    }
    for (int i=fir[u];i;i=ne[i])
        if ((v=to[i])!=fa[u]&&v!=son[u])
        {
            top[v]=v;
            dfs2(v);
        }
}
void build(int p,int L,int R)
{
    val[p]=-;
    if (L==R) return;
    int mid=L+R>>;
    if (L<=mid) build(p*2,L,mid);
    if (mid<R) build(p*2+,mid+,R);
}
void init()
{
    int u,v;
    scanf("%d%d",&n,&q);
    for (int i=;i<n;i++)
    {
        scanf("%d%d",&u,&v);
        add(i*2,u,v);
        add(i*2+,v,u);
    }
    dep[]=;
    dfs1();
    top[]=;
    dfs2();
    build(,,n);
}
void modi(int p,int L,int R,int u)
{
    int x,y;
    if (L==R)
    {
        if (val[p]==-) val[p]=u;
        else val[p]=-;
        return;
    }
    int mid=L+R>>;
    if (pos[u]<=mid) modi(p*2,L,mid,u);
    else modi(p*2+,mid+,R,u);
    x=L<=mid?val[p*2]:-;
    y=mid<R?val[p*2+]:-;
    if (x==-||(y>&&dep[y]<dep[x])) val[p]=y;
    else val[p]=x;
}
int qry(int p,int L,int R,int l,int r)
{
    int x,y;
    if (l<=L&&R<=r) return val[p];
    int mid=L+R>>;
    x=l<=mid?qry(p*2,L,mid,l,r):-;
    y=r>mid?qry(p*2+,mid+,R,l,r):-;
    if (x==-||(y>&&dep[y]<dep[x])) return y;
    return x;
}
int query(int u)
{
    int ans=-,x;
    while (top[u]!=)
    {
        x=qry(,,n,pos[top[u]],pos[u]);
        if (x>&&(ans==-||dep[x]<dep[ans])) ans=x;
        u=fa[top[u]];
    }
    x=qry(,,n,,pos[u]);
    if (x>&&(ans==-||dep[x]<dep[ans])) ans=x;
    return ans;
}
int main()
{
    int u,x;
    init();
    while (q--)
    {
        scanf("%d%d",&x,&u);
        if (x) printf("%d\n",query(u));
        else modi(,,n,u);
    }
}