Count on a tree SPOJ

You are given a tree with N nodes.The tree nodes are numbered from 1 to N.Each node has an integer weight.

We will ask you to perform the following operation:

u v k : ask for the kth minimum weight on the path from node u to node v

Input

In the first line there are two integers N and M.(N,M<=100000)

In the second line there are N integers.The ith integer denotes the weight of the ith node.

In the next N-1 lines,each line contains two integers u v,which describes an edge (u,v).

In the next M lines,each line contains three integers u v k,which means an operation asking for the kth minimum weight on the path from node u to node v.

Output

For each operation,print its result.

Example

Input:

8 5

8 5

105 2 9 3 8 5 7 7
1 2
1 3
1 4
3 5
3 6
3 7
4 8
2 5 1
2 5 2
2 5 3
2 5 4
7 8 2
Output:
2
8
9
105
7
题意：给你一棵由n个节点，n-1条边组成的树。问u，v之间第k小的是什么。
题解：第k小，首先想到主席树，我们建一棵由根开始的线段树。则u，v之间的关系就可以表示成 root[u]+root[v]-root[lca(u,v)]-root[fa[lca(u,v)]].因为需要知道lca，所以再套一个lca。
代码：

#include<bits/stdc++.h>using namespace std;const int N=1e6+10;int cnt,u,v,k,tot,n,m,mx;int dep[N],fa[N],siz[N],son[N];int a[N],root[N];struct node{int to,next;}edge[N<<2];struct nodee{int l,r,sum;}T[N*40];int head[N<<2];vector<int>p;int getid(int x){return lower_bound(p.begin(),p.end(),x)-p.begin()+1;}void add(int u,int v){    edge[cnt].to=v,edge[cnt].next=head[u],head[u]=cnt++;    edge[cnt].to=u,edge[cnt].next=head[v],head[v]=cnt++;}void update(int l,int r,int &x,int y,int pos){    T[++cnt]=T[y],x=cnt,T[x].sum++;    if(l==r) return ;    int mid=(r+l)>>1;    if(pos<=mid)        update(l,mid,T[x].l,T[y].l,pos);    else        update(mid+1,r,T[x].r,T[y].r,pos);}int query(int l,int r,int x,int y,int z,int rr,int k){    if(l>=r) return l;    int mid=(r+l)>>1;    int sum=T[T[y].l].sum+T[T[x].l].sum-T[T[z].l].sum-T[T[rr].l].sum;    if(k<=sum) query(l,mid,T[x].l,T[y].l,T[z].l,T[rr].l,k);    else query(mid+1,r,T[x].r,T[y].r,T[z].r,T[rr].r,k-sum);}void dfs1(int u,int f,int d){    dep[u]=d,fa[u]=f,siz[u]=1,son[u]=0;    for(int i=head[u];i!=-1;i=edge[i].next)    {        int v=edge[i].to;        if(v==f)continue;        dfs1(v,u,d+1);        siz[u]+=siz[v];        if(siz[son[u]]<siz[v]) son[u]=v;    }}int top[N],pre[N],tree[N];void dfs2(int u,int tp){    top[u]=tp,tree[u]=++tot,pre[tree[u]]=u;    update(1,mx,root[u],root[fa[u]],getid(a[u]));    if(!son[u]) return ;    dfs2(son[u],tp);    for(int i=head[u];i!=-1;i=edge[i].next)    {        int v=edge[i].to;        if(v==fa[u]||v==son[u]) continue;        dfs2(v,v);    }}int lca(int x,int y){        int fx=top[x],fy=top[y];        while(fx!=fy)        {            if(dep[fx]<dep[fy]) swap(x,y),swap(fx,fy);            x=fa[fx],fx=top[x];        }        if(dep[x]>dep[y]) swap(x,y);        return x;}void init(){    tot=0;    cnt=0;    memset(head,-1,sizeof(head));}int main(){    init();    scanf("%d%d",&n,&m);    for(int i=1;i<=n;i++)    {        scanf("%d",&a[i]);        p.push_back(a[i]);    }    for(int i=1;i<n;i++)    {        scanf("%d%d",&u,&v);        add(u,v);    }    cnt=0;    sort(p.begin(),p.end()),p.erase(unique(p.begin(),p.end()),p.end());    mx=p.size();    dfs1(1,0,1),dfs2(1,1);    while(m--)    {        scanf("%d%d%d",&u,&v,&k);        int lc=lca(u,v);        printf("%d\n",p[query(1,mx,root[u],root[v],root[lc],root[fa[lc]],k)-1]);    }}