【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=0,f=1;char ch=getchar();
    while((ch<'0'||ch>'9')&&ch!='-')ch=getchar();
    if(ch=='-')f=-1;
    while(ch<='9'&&ch>='0')x=x*10+ch-'0',ch=getchar();
    return f*x;
}

struct ed{
    int to,last;
}edge[maxn*2];

struct lc{
    int father,num[11];
}f[19][maxn];

int last[maxn],tot=0,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!=-1)f[0][poi].father=lastt;
    for(int i=last[poi];i;i=edge[i].last)
        if(edge[i].to!=lastt)dfs(edge[i].to,poi,de+1);
}

void update(int pos,int cen)
{
    int i=1,j=1;
    while(i<=f[cen-1][f[cen-1][pos].father].num[0]&&j<=f[cen-1][pos].num[0]&&f[cen][pos].num[0]<10)
    {
        if(i<=f[cen-1][f[cen-1][pos].father].num[0]&&f[cen-1][f[cen-1][pos].father].num[i]<f[cen-1][pos].num[j])f[cen][pos].num[++f[cen][pos].num[0]]=f[cen-1][f[cen-1][pos].father].num[i++];
        else if(j<=f[cen-1][pos].num[0])f[cen][pos].num[++f[cen][pos].num[0]]=f[cen-1][pos].num[j++];
    }
    while(i<=f[cen-1][f[cen-1][pos].father].num[0]&&f[cen][pos].num[0]<10)f[cen][pos].num[++f[cen][pos].num[0]]=f[cen-1][f[cen-1][pos].father].num[i++];
    while(j<=f[cen-1][pos].num[0]&&f[cen][pos].num[0]<10)f[cen][pos].num[++f[cen][pos].num[0]]=f[cen-1][pos].num[j++];
}

void pre()
{
    for(int i=1;(1<<i)<=n;i++)
        for(int j=1;j<=n;j++)
            if(f[i-1][f[i-1][j].father].father)f[i][j].father=f[i-1][f[i-1][j].father].father,update(j,i);
}

int ANS[11],ANS_B[11];

void Up(int pos,int cen,int kk)
{
    ANS_B[0]=0;
    int i=1,j=1;
    while(i<=ANS[0]&&j<=f[cen][pos].num[0]&&ANS_B[0]<kk)
    {
        if(i<=ANS[0]&&ANS[i]<f[cen][pos].num[j])ANS_B[++ANS_B[0]]=ANS[i++];
        else if(j<=f[cen][pos].num[0])ANS_B[++ANS_B[0]]=f[cen][pos].num[j++];
    }
    while(i<=ANS[0]&&ANS_B[0]<kk)ANS_B[++ANS_B[0]]=ANS[i++];
    while(j<=f[cen][pos].num[0]&&ANS_B[0]<kk)ANS_B[++ANS_B[0]]=f[cen][pos].num[j++];
    for(int i=0;i<=ANS_B[0];i++)ANS[i]=ANS_B[i];
}

void print()
{
    printf("%d",ANS[0]);
    for(int i=1;i<=ANS[0];i++)printf(" %d",ANS[i]);
    printf("
");
}

void lca(int u,int v,int kk)
{
    ANS[0]=0;
    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>=0;i--)
            if((1<<i) & (dep[v]-dep[u]))Up(v,i,kk),v=f[i][v].father;
    }
    if(u==v)
    {
        Up(u,0,kk);
        print();
        return;
    }
    nu=log2(dep[v]);
    while(nu!=-1)
    {
        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,0,kk);
    Up(u,0,kk);
    Up(f[0][u].father,0,kk);
    print();
}

int main()
{
    freopen("t.in","r",stdin);
    int m,q,u,v,kk;
    n=in(),m=in(),q=in();
    for(int i=1;i<n;i++)
        u=in(),v=in(),add(u,v);
    for(int i=1;i<=m;i++)
    {
        u=in();
        if(f[0][u].num[0]<10)f[0][u].num[++f[0][u].num[0]]=i;
    }
    dfs(1,-1,0);
    pre();
    for(int i=1;i<=q;i++)
    {
        u=in(),v=in(),kk=in();
        lca(u,v,kk);
    }
    return 0;
}