spoj COT

COT - Count on a tree

#tree

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 5105 2 9 3 8 5 7 71 2        1 31 43 53 63 74 8
2 5 1
2 5 2
2 5 3
2 5 4
7 8 2 

Output:2
8
9
105
7 

解：第k大所在区间为sum[l]+sum[r]-sum[lca(l,r)]-sum[fa[lca(l,r)]]

#include<iostream>#include<algorithm>#include<cstdio>#include<cstdlib>#include<cstring>#include<vector>#include<map>#include <bits/stdc++.h>using namespace std;const int N = 1e5+100;typedef long long LL;int rt[N*20], ls[N*20], rs[N*20], sum[N*20];int fa[2*N][30], dep[2*N], vis[N];int a[N], b[N], tot, cnt, head[N], len;struct node{    int to, next;} p[2*N];void init(){    memset(head,-1,sizeof(head));    memset(vis,0,sizeof(vis));    cnt=0;    return ;}void add(int u,int v){    p[cnt].to=v,p[cnt].next=head[u];head[u]=cnt++;    p[cnt].to=u,p[cnt].next=head[v];head[v]=cnt++;    return ;}void build(int &o,int l,int r){    o= ++tot,sum[o]=0;    if(l==r) return ;    int mid=(l+r)/2;    build(ls[o],l,mid);    build(rs[o],mid+1,r);    return ;}void update(int &o,int l,int r,int last,int p){    o= ++tot;    ls[o]=ls[last],rs[o]=rs[last];    sum[o]=sum[last]+1;    if(l==r) return ;    int mid=(l+r)/2;    if(p<=mid) update(ls[o],l,mid,ls[last],p);    else update(rs[o],mid+1,r,rs[last],p);    return ;}int query(int ss,int tt,int s1,int t1,int l,int r,int cnt){    if(l==r) return l;    int tmp=sum[ls[tt]]+sum[ls[ss]]-sum[ls[s1]]-sum[ls[t1]];    int mid=(l+r)/2;    if(tmp>=cnt) return query(ls[ss],ls[tt],ls[s1],ls[t1],l,mid,cnt);    else return query(rs[ss],rs[tt],rs[s1],rs[t1],mid+1,r,cnt-tmp);}void dfs(int u,int d,int f,int root){    vis[u]=1,dep[u]=d,fa[u][0]=f;    update(rt[u],1,len,root,a[u]);    root=rt[u];    for(int i=head[u];i!=-1;i=p[i].next)    {        int v=p[i].to;        if(vis[v]) continue;        dfs(v,d+1,u,root);    }    return ;}void lca(int n){    int k=(int)(log(1.0*n)/log(2.0));    for(int i=1;i<=k;i++)    {        for(int j=1;j<=n;j++)        {            fa[j][i]=fa[fa[j][i-1]][i-1];        }    }    return ;}int get(int x,int y,int n){    if(dep[x]<dep[y]) swap(x,y);    int k=(int)(log(1.0*n)/log(2.0));    int d=dep[x]-dep[y];    for(int i=0;i<=k;i++)        if((d&(1<<i))) x=fa[x][i];    if(x==y) return x;    for(int i=k;i>=0;i--)    {        if(fa[x][i]!=fa[y][i]) x=fa[x][i],y=fa[y][i];    }    return fa[x][0];}int main(){    int t, n, q;    scanf("%d %d", &n, &q);    for(int i=1; i<=n; i++) scanf("%d", &a[i]), b[i]=a[i];    sort(b+1,b+n+1);    len=unique(b+1,b+n+1)-(b+1);    tot=0;    build(rt[0],1,len);    for(int i=1; i<=n; i++)  a[i]=lower_bound(b+1,b+len+1,a[i])-(b);    init();    for(int i=0;i<n-1;i++)    {        int x, y;        scanf("%d %d", &x, &y);        add(x,y);    }    dfs(1,1,0,rt[0]);    lca(n);    while(q--)    {        int l, r, x;        scanf("%d %d %d", &l, &r, &x);        int pos=get(l,r,n);        printf("%d\n",b[query(rt[l],rt[r],rt[pos],rt[fa[pos][0]],1,len,x)]);    }    return 0;}