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

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.

$\text{[math]}$

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

output

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。。

  1 #include<cstdio>
  2 #include<cstring>
  3 #include<algorithm>
  4 #include<cmath>
  5 #include<stack>
  6 #include<vector>
  7 
  8 #define maxn 100001
  9 
 10 using namespace std;
 11 
 12 inline int in()
 13 {
 14     int x=0,f=1;char ch=getchar();
 15     while((ch<'0'||ch>'9')&&ch!='-')ch=getchar();
 16     if(ch=='-')f=-1;
 17     while(ch<='9'&&ch>='0')x=x*10+ch-'0',ch=getchar();
 18     return f*x;
 19 }
 20 
 21 struct ed{
 22     int to,last;
 23 }edge[maxn*2];
 24 
 25 struct lc{
 26     int father,num[11];
 27 }f[19][maxn];
 28 
 29 int last[maxn],tot=0,dep[maxn],n;
 30 
 31 void add(int u,int v)
 32 {
 33     edge[++tot].to=v,edge[tot].last=last[u],last[u]=tot;
 34     edge[++tot].to=u,edge[tot].last=last[v],last[v]=tot;
 35 }
 36 
 37 void dfs(int poi,int lastt,int de)
 38 {
 39     dep[poi]=de;
 40     if(lastt!=-1)f[0][poi].father=lastt;
 41     for(int i=last[poi];i;i=edge[i].last)
 42         if(edge[i].to!=lastt)dfs(edge[i].to,poi,de+1);
 43 }
 44 
 45 void update(int pos,int cen)
 46 {
 47     int i=1,j=1;
 48     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)
 49     {
 50         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++];
 51         else if(j<=f[cen-1][pos].num[0])f[cen][pos].num[++f[cen][pos].num[0]]=f[cen-1][pos].num[j++];
 52     }
 53     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++];
 54     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++];
 55 }
 56 
 57 void pre()
 58 {
 59     for(int i=1;(1<<i)<=n;i++)
 60         for(int j=1;j<=n;j++)
 61             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);
 62 }
 63 
 64 int ANS[11],ANS_B[11];
 65 
 66 void Up(int pos,int cen,int kk)
 67 {
 68     ANS_B[0]=0;
 69     int i=1,j=1;
 70     while(i<=ANS[0]&&j<=f[cen][pos].num[0]&&ANS_B[0]<kk)
 71     {
 72         if(i<=ANS[0]&&ANS[i]<f[cen][pos].num[j])ANS_B[++ANS_B[0]]=ANS[i++];
 73         else if(j<=f[cen][pos].num[0])ANS_B[++ANS_B[0]]=f[cen][pos].num[j++];
 74     }
 75     while(i<=ANS[0]&&ANS_B[0]<kk)ANS_B[++ANS_B[0]]=ANS[i++];
 76     while(j<=f[cen][pos].num[0]&&ANS_B[0]<kk)ANS_B[++ANS_B[0]]=f[cen][pos].num[j++];
 77     for(int i=0;i<=ANS_B[0];i++)ANS[i]=ANS_B[i];
 78 }
 79 
 80 void print()
 81 {
 82     printf("%d",ANS[0]);
 83     for(int i=1;i<=ANS[0];i++)printf(" %d",ANS[i]);
 84     printf("
");
 85 }
 86 
 87 void lca(int u,int v,int kk)
 88 {
 89     ANS[0]=0;
 90     int nu;
 91     if(dep[u]>dep[v])swap(u,v);
 92     if(dep[v]!=dep[u])
 93     {    
 94         nu=log2(dep[v]-dep[u]);
 95         for(int i=nu;i>=0;i--)
 96             if((1<<i) & (dep[v]-dep[u]))Up(v,i,kk),v=f[i][v].father;
 97     }
 98     if(u==v)
 99     {
100         Up(u,0,kk);
101         print();
102         return;
103     }
104     nu=log2(dep[v]);
105     while(nu!=-1)
106     {
107         if(f[nu][u].father==f[nu][v].father){nu--;continue;}
108         Up(v,nu,kk);
109         Up(u,nu,kk);
110         u=f[nu][u].father;
111         v=f[nu][v].father;
112         nu--;
113     }
114     Up(v,0,kk);
115     Up(u,0,kk);
116     Up(f[0][u].father,0,kk);
117     print();
118 }
119 
120 int main()
121 {
122     freopen("t.in","r",stdin);
123     int m,q,u,v,kk;
124     n=in(),m=in(),q=in();
125     for(int i=1;i<n;i++)
126         u=in(),v=in(),add(u,v);
127     for(int i=1;i<=m;i++)
128     {
129         u=in();
130         if(f[0][u].num[0]<10)f[0][u].num[++f[0][u].num[0]]=i;
131     }
132     dfs(1,-1,0);
133     pre();
134     for(int i=1;i<=q;i++)
135     {
136         u=in(),v=in(),kk=in();
137         lca(u,v,kk);
138     }
139     return 0;
140 }

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