P2633 Count on a tree tree tree Chairman

Title Description

Given a tree of N nodes, each point has a weight, for M interrogation (u, v, k), you need to answer the K dots smaller weights v and u xor lastans between these two nodes. Lastans which is one answer to the inquiry, initially 0, which is the first inquiry is expressly u.

Input Format

The first line of two integers N, M.

The second line has N integers, wherein the integer represents the i-th point i of the weight.

Behind the N-1 lines each two integers (x, y), the point x to point y indicates there is an edge.

The last two lines each integers M (u, v, k), represents a group of query.

Output Format

M line represents the answer to each inquiry.

Sample input and output

Input # 1copy

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
0 5 2
10 5 3
11 5 4
110 8 2

Output # 1copy

2 
. 8 
. 9 
105 
. 7 


tree of k small 


Note 
Note President tree query x = t [lson [L] ] + t [lson [R]] - t [lson [lca]] - t [lson [falca]];

#include<bits/stdc++.h>
using namespace std;
//input by bxd
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define repp(i,a,b) for(int i=(a);i>=(b);--i)
#define RI(n) scanf("%d",&(n))
#define RII(n,m) scanf("%d%d",&n,&m)
#define RIII(n,m,k) scanf("%d%d%d",&n,&m,&k)
#define RS(s) scanf("%s",s)
#define ll long long
#define see(x) (cerr<<(#x)<<'='<<(x)<<endl)
#define pb push_back
#define inf 0x3f3f3f3f
#define pb push_back
#define CLR(A,v)  memset(A,v,sizeof A)

typedef pair<int,int>pii;
//////////////////////////////////
const int N=1e5+5;

int a[N],b[N],T[N],lson[N<<5],rson[N<<5],t[N<<5],tot,c[N];

void upnode(int pre,int x,int l,int r,int &pos)
{
    pos=++tot;
    lson[pos]=lson[pre],rson[pos]=rson[pre],t[pos]=t[pre]+1;
    if(l==r)return ;int m=(l+r)>>1;
    if(x<=m)upnode(lson[pre],x,l,m,lson[pos]);
    else upnode(rson[pre],x,m+1,r,rson[pos]);
}
int qsum(int L,int R,int lca,int falca,int k,int l,int r)
{
    if(l==r)return l;
    int x=t[lson[R]]+t[lson[L]]-t[lson[lca]]-t[lson[falca]];
    int m=(l+r)>>1;
    if(x>=k)return qsum(lson[L],lson[R],lson[lca],lson[falca],k, l,m);
    else return qsum(rson[L],rson[R],rson[lca],rson[falca],k-x, m+1,r);
}
///////////////////////////
int head[N],pos,id[N],top[N],fa[N],son[N],siz[N],dep[N],cnt;
int n,m;
struct Edge{int to,nex; } edge[N<<2];
inline void add(int a,int b)
{
    edge[++pos]=(Edge){b,head[a]};
    head[a]=pos;
}
void dfs1(int x,int f)
{
    fa[x]=f;son[x]=0;siz[x]=1;dep[x]=dep[f]+1;
    upnode(T[f],c[x],1,n,T[x]);
    for(int i=head[x];i;i=edge[i].nex)
    {
        int v=edge[i].to;if(v==f)continue;
        dfs1(v,x);siz[x]+=siz[v];
        if(siz[son[x]]<siz[v])son[x]=v;
    }
}
void dfs2(int x,int topf)
{
    top[x]=topf;
    if(!son[x])return ;
    dfs2(son[x],topf);
    for(int i=head[x];i;i=edge[i].nex)
    {
        int v=edge[i].to;
        if(v==fa[x]||v==son[x])continue;
        dfs2(v,v);
    }
}
int getlca(int x,int y)
{
    while(top[x]!=top[y])
    {
        if(dep[top[x]]<dep[top[y]])swap(x,y);
        x=fa[top[x]];
    }
    return dep[x]>=dep[y]?y:x;
}
int q;
int main()
{
    RII(n,q);
    rep(i,1,n)RI(a[i]),b[i]=a[i];
    sort(b+1,b+1+n);
    int m=unique(b+1,b+1+n)-b-1;
    rep(i,1,n)
    c[i]=lower_bound(b+1,b+1+m,a[i])-b;

    rep(i,1,n-1)
    {
        int u,v;RII(u,v);add(u,v);add(v,u);
    }
    dfs1(1,0);dfs2(1,1);
    int ans=0,lca;
    while(q--)
    {
        int x,y,k;RIII(x,y,k);
        x^=ans,lca=getlca(x,y);
        ans=b[ qsum(T[x],T[y],T[lca],T[fa[lca]],k,1,n)];
        printf("%d\n",ans);
    }
    return 0;
}

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