【Codeforces Round 326 (Div 2)E】【树链剖分】Duff in the Army 树上给定路径上编号最小的几个人

E. Duff in the Army

time limit per test

4 seconds

memory limit per test

512 megabytes

input

standard input

output

standard output

Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is iand he/she lives in city number c_i. Note that there may be more than one person in a city, also there may be no people living in the city.

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p₁, p₂, ..., p_x in increasing order.

If k = min(x, a), then Duff should tell Malek numbers k, p₁, p₂, ..., p_k in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 10⁵).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c₁, c₂, ..., c_m separated by spaces (1 ≤ c_i ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p₁, p₂, ..., p_k separated by spaces in one line.

Sample test(s)

input

5 4 51 31 21 44 52 1 4 34 5 61 5 25 5 102 3 35 3 1

output

1 32 2 303 1 2 41 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

#include<stdio.h>#include<string.h>#include<ctype.h>#include<math.h>#include<iostream>#include<string>#include<set>#include<map>#include<vector>#include<queue>#include<bitset>#include<algorithm>#include<time.h>using namespace std;void fre(){freopen("c://test//input.in","r",stdin);freopen("c://test//output.out","w",stdout);}#define MS(x,y) memset(x,y,sizeof(x))#define MC(x,y) memcpy(x,y,sizeof(x))#define MP(x,y) make_pair(x,y)#define ls o<<1#define rs o<<1|1typedef long long LL;typedef unsigned long long UL;typedef unsigned int UI;template <class T1,class T2>inline void gmax(T1 &a,T2 b){if(b>a)a=b;}template <class T1,class T2>inline void gmin(T1 &a,T2 b){if(b<a)a=b;}const int N=1e5+10,M=0,Z=1e9+7,ms63=1061109567;int casenum,casei;int n,m,q,tim;int x,y,w;struct A{int l,r;int g;int b[10];}c[1<<18];vector<int>a[N];int g[N];int b[N][10];int G,tmpg;int B[10],tmpb[10];int fa[N],son[N],dep[N],size[N],pos[N],rev[N],top[N];void dfs1(int x){size[x]=1;son[x]=0;for(int i=a[x].size()-1;~i;--i){int y=a[x][i];if(y==fa[x])continue;fa[y]=x;dep[y]=dep[x]+1;dfs1(y);size[x]+=size[y];if(size[y]>size[son[x]])son[x]=y;}}void dfs2(int x,int chain){pos[x]=++tim;rev[tim]=x;top[x]=chain;if(son[x]==0)return;dfs2(son[x],chain);for(int i=a[x].size()-1;~i;--i){int y=a[x][i];if(y!=fa[x]&&y!=son[x])dfs2(y,y);}}void merge(int &g,int &gl,int &gr,int b[],int bl[],int br[]){int p=0,pl=0,pr=0;while(p<10){if(pl<gl&&pr<gr){if(bl[pl]<br[pr])b[p++]=bl[pl++];else b[p++]=br[pr++];}else if(pl<gl)b[p++]=bl[pl++];else if(pr<gr)b[p++]=br[pr++];else break;}g=p;}void pushup(int o){merge(c[o].g,c[ls].g,c[rs].g,c[o].b,c[ls].b,c[rs].b);}void build(int o,int l,int r){c[o].l=l;c[o].r=r;if(l==r){int v=rev[l];c[o].g=g[v];MC(c[o].b,b[v]);return;}int m=(l+r)>>1;build(ls,l,m);build(rs,m+1,r);pushup(o);}void update(int o,int l,int r){if(c[o].l==l&&c[o].r==r){tmpg=G;MC(tmpb,B);merge(G,tmpg,c[o].g,B,tmpb,c[o].b);return;}int m=(c[o].l+c[o].r)>>1;if(r<=m)update(ls,l,r);else if(l>m)update(rs,l,r);else{update(ls,l,m);update(rs,m+1,r);}}void SOLVE(int x,int y,int w){G=0;while(top[x]!=top[y]){if(dep[top[x]]<dep[top[y]])swap(x,y);update(1,pos[top[x]],pos[x]);x=fa[top[x]];}if(pos[x]>pos[y])swap(x,y);update(1,pos[x],pos[y]);gmin(G,w);printf("%d",G);for(int i=0;i<G;i++)printf(" %d",B[i]);puts("");}int main(){while(~scanf("%d%d%d",&n,&m,&q)){for(int i=1;i<=n;i++){a[i].clear();g[i]=0;}for(int i=1;i<n;i++){scanf("%d%d",&x,&y);a[x].push_back(y);a[y].push_back(x);}for(int i=1;i<=m;i++){scanf("%d",&x);if(g[x]<10)b[x][g[x]++]=i;}dep[1]=0;fa[1]=0;dfs1(1);tim=0;dfs2(1,1);build(1,1,n);for(int i=1;i<=q;i++){scanf("%d%d%d",&x,&y,&w);SOLVE(x,y,w);}}return 0;}/*【trick&&吐槽】这题明明要比D简单，是裸的树链剖分。我写得很慢慢悠悠，然而竟然是div2前几个做出来的，太吃惊了！【题意】给你一棵有n（1e5）个节点树，然后有m（1e5）个人，告诉你编号为i个人位于什么节点。有q（1e5）个询问，问你对于（x,y）路径上的点中，最小编号的w（[1,10]）个人的编号。（当然如果没有w个人，有几个输出几个）【类型】树链剖分【分析】因为w实在是太小了，所以每个节点直接记录以它为父节点的最小编号的10个人的编号即可。用线段树，通过类似于归并排序的方式维护一下。然后因为是在树上，套一个树链剖分的模板就可以AC了。因为没有修改，所以线段树只用pushup，也就写个合并操作即可。【时间复杂度&&优化】O（（n+q）logn*10）*/