【LCA】CodeForce #326 Div.2 E:Duff in the Army

C. Duff in the Army

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 ci. 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, ..., px in increasing order.

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk 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, ..., cm separated by spaces (1 ≤ ci ≤ 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, ..., pk separated by spaces in one line.

Sample test(s)

input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1

output

1 3
2 2 3
0
3 1 2 4
1 2

Note

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

---

　　大约题目是给一棵树给m个人在哪个点上的信息

　　然后给q个询问，每次问u到v上的路径有的点上编号最小的k个人，k<=10（很关键）

　　u到v上路径的询问很容易想到lca

　　但是前k个答案很不好搞？

　　直接在lca数组里面开个Num[11]记录前10个在该点上的编号

　　码了1个半小时结果wa成狗- -

　　最后发现lca打挂了。。

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<cmath>
 #include<stack>
 #include<vector>

 #define maxn 100001

 using namespace std;

 inline int in()
 {
     int x=,f=;char ch=getchar();
     while((ch<''||ch>'')&&ch!='-')ch=getchar();
     if(ch=='-')f=-;
     while(ch<=''&&ch>='')x=x*+ch-'',ch=getchar();
     return f*x;
 }

 struct ed{
     int to,last;
 }edge[maxn*];

 struct lc{
     int father,num[];
 }f[][maxn];

 int last[maxn],tot=,dep[maxn],n;

 void add(int u,int v)
 {
     edge[++tot].to=v,edge[tot].last=last[u],last[u]=tot;
     edge[++tot].to=u,edge[tot].last=last[v],last[v]=tot;
 }

 void dfs(int poi,int lastt,int de)
 {
     dep[poi]=de;
     if(lastt!=-)f[][poi].father=lastt;
     for(int i=last[poi];i;i=edge[i].last)
         if(edge[i].to!=lastt)dfs(edge[i].to,poi,de+);
 }

 void update(int pos,int cen)
 {
     int i=,j=;
     while(i<=f[cen-][f[cen-][pos].father].num[]&&j<=f[cen-][pos].num[]&&f[cen][pos].num[]<)
     {
         if(i<=f[cen-][f[cen-][pos].father].num[]&&f[cen-][f[cen-][pos].father].num[i]<f[cen-][pos].num[j])f[cen][pos].num[++f[cen][pos].num[]]=f[cen-][f[cen-][pos].father].num[i++];
         else if(j<=f[cen-][pos].num[])f[cen][pos].num[++f[cen][pos].num[]]=f[cen-][pos].num[j++];
     }
     while(i<=f[cen-][f[cen-][pos].father].num[]&&f[cen][pos].num[]<)f[cen][pos].num[++f[cen][pos].num[]]=f[cen-][f[cen-][pos].father].num[i++];
     while(j<=f[cen-][pos].num[]&&f[cen][pos].num[]<)f[cen][pos].num[++f[cen][pos].num[]]=f[cen-][pos].num[j++];
 }

 void pre()
 {
     for(int i=;(<<i)<=n;i++)
         for(int j=;j<=n;j++)
             if(f[i-][f[i-][j].father].father)f[i][j].father=f[i-][f[i-][j].father].father,update(j,i);
 }

 int ANS[],ANS_B[];

 void Up(int pos,int cen,int kk)
 {
     ANS_B[]=;
     int i=,j=;
     while(i<=ANS[]&&j<=f[cen][pos].num[]&&ANS_B[]<kk)
     {
         if(i<=ANS[]&&ANS[i]<f[cen][pos].num[j])ANS_B[++ANS_B[]]=ANS[i++];
         else if(j<=f[cen][pos].num[])ANS_B[++ANS_B[]]=f[cen][pos].num[j++];
     }
     while(i<=ANS[]&&ANS_B[]<kk)ANS_B[++ANS_B[]]=ANS[i++];
     while(j<=f[cen][pos].num[]&&ANS_B[]<kk)ANS_B[++ANS_B[]]=f[cen][pos].num[j++];
     for(int i=;i<=ANS_B[];i++)ANS[i]=ANS_B[i];
 }

 void print()
 {
     printf("%d",ANS[]);
     for(int i=;i<=ANS[];i++)printf(" %d",ANS[i]);
     printf("\n");
 }

 void lca(int u,int v,int kk)
 {
     ANS[]=;
     int nu;
     if(dep[u]>dep[v])swap(u,v);
     if(dep[v]!=dep[u])
     {
         nu=log2(dep[v]-dep[u]);
         for(int i=nu;i>=;i--)
             if((<<i) & (dep[v]-dep[u]))Up(v,i,kk),v=f[i][v].father;
     }
     if(u==v)
     {
         Up(u,,kk);
         print();
         return;
     }
     nu=log2(dep[v]);
     while(nu!=-)
     {
         if(f[nu][u].father==f[nu][v].father){nu--;continue;}
         Up(v,nu,kk);
         Up(u,nu,kk);
         u=f[nu][u].father;
         v=f[nu][v].father;
         nu--;
     }
     Up(v,,kk);
     Up(u,,kk);
     Up(f[][u].father,,kk);
     print();
 }

 int main()
 {
     freopen("t.in","r",stdin);
     int m,q,u,v,kk;
     n=in(),m=in(),q=in();
     for(int i=;i<n;i++)
         u=in(),v=in(),add(u,v);
     for(int i=;i<=m;i++)
     {
         u=in();
         if(f[][u].num[]<)f[][u].num[++f[][u].num[]]=i;
     }
     dfs(,-,);
     pre();
     for(int i=;i<=q;i++)
     {
         u=in(),v=in(),kk=in();
         lca(u,v,kk);
     }
     return ;
 }