# 【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.

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 ≤ nv ≠ 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 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):

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
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("\n");
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++)
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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posted @ 2015-10-16 17:41  puck_just_me  阅读(...)  评论(... 编辑 收藏