codeforces 587C(树上的倍增算法)

分析：

同LCA相似的思路，维护st[ i ][ j ] 代表从节点i往上（不包含i）走2^j个节点前十个最小值集合，这样一直倍增着往上走秩序logn次集合合并即可。

#include <bits/stdc++.h>

#define fst first
#define snd second
#define ALL(a) a.begin(), a.end()
#define clr(a, x) memset(a, x, sizeof a)
#define rep(i,x) for(int i=0;i<(int)x;i++)
#define rep1(i,x,y) for(int i=x;i<=(int)y;i++)
#define LOGN 18
typedef long long ll;
using namespace std;

const int N = 1e5 + 100;
vector<int> G
, pre
;
int posi
,depth
,fa
[LOGN],n;
vector<int> st
[LOGN];
void get(vector<int>& a , vector<int>& b){
for(auto v : a) b.push_back(v);
sort(ALL(b)); while(b.size() > 10) b.pop_back();
}
void dfs(int u,int f,int dep){
depth[u] = dep; fa[u][0] = f;
if(f != -1) get(pre[f], st[u][0]);
for(auto v:G[u])
if(v != f) dfs( v , u , dep+1);
}
void build(){
for(int i=0;i<LOGN-1;i++)
for(int j=1;j<=n;j++){
fa[j][i+1] = (fa[j][i]==-1 ? -1 : fa[fa[j][i]][i]);
if(fa[j][i+1] != -1) {
get(st[j][i] , st[j][i+1]);
get(st[fa[j][i]][i], st[j][i+1]);
}
}
}
void show(int a,vector<int>& ans){
int lim = min(a , (int)ans.size());
printf("%d",lim);
rep(i,lim) printf(" %d",ans[i]);
printf("\n");
}
int cal(int u,int v ,int a){
if(depth[u] > depth[v]) swap(u , v);
int cha = depth[v] - depth[u];
vector<int> ans;
get(pre[v] , ans);
for(int i=LOGN - 1; i>=0;i--){
if((cha>>i)&1){
get(st[v][i] , ans);
v = fa[v][i];
}
}
if(u == v){
show(a , ans);
return u;
}
get(pre[u] , ans);
for(int i=LOGN - 1; i>=0;i--){
if(fa[v][i] != fa[u][i]){
get(st[v][i] ,ans);
get(st[u][i] ,ans);
v = fa[v][i];
u = fa[u][i];
}
}
get(st[u][0] , ans);
show(a , ans);
return fa[u][0];
}
int m,q;
void read(){
scanf("%d %d %d",&n,&m,&q);
rep(i,n-1){
int u, v;
scanf("%d %d",&u,&v);
G[u].push_back(v);
G[v].push_back(u);
}
rep1(i,1,m) scanf("%d",&posi[i]),pre[posi[i]].push_back(i);
dfs(1 , -1 , 0);
build();
rep1(i,1,n)if(pre[i].size()){
sort(ALL(pre[i]));
while(pre[i].size() > 10) pre[i].pop_back();
}
int u , v , a;
while(q--){
scanf("%d %d %d",&u,&v,&a);
cal(u , v , a);
}
}
int main()
{
read();
return 0;
}