CodeForces 587C :Duff in the Arm 主席树 + 树上倍增 + LCA

传送门

题意

分析

两种思路
第一种是从根节点向下建立主席树，在$dfs$的过程中插入并维护$fa[u][i]$数组求$LCA$，然后通过四个点$u,v,LCA(u,v),fa[LCA(u,v)][0]$来确定这个区间内出现了多少个数，然后枚举前$k$小个数即可
第二种思路留个坑以后补

代码

#pragma GCC optimize(3)
#include <bits/stdc++.h>
#define debug(x) cout<<#x<<":"<<x<<endl;
#define dl(x) printf("%lld\n",x);
#define di(x) printf("%d\n",x);
#define _CRT_SECURE_NO_WARNINGS
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
typedef vector<int> VI;
const int INF = 0x3f3f3f3f;
const int N = 2e5 + 10,M = N * 2;
const ll mod = 1000000007;
const double eps = 1e-9;
const double PI = acos(-1);
template<typename T>inline void read(T &a) {
    char c = getchar(); T x = 0, f = 1; while (!isdigit(c)) {if (c == '-')f = -1; c = getchar();}
    while (isdigit(c)) {x = (x << 1) + (x << 3) + c - '0'; c = getchar();} a = f * x;
}
int gcd(int a, int b) {return (b > 0) ? gcd(b, a % b) : a;}
int h[N],ne[M],e[M],num;
int f[N][20],dep[N];
int n,m,q,idx;
int root[N];
VI id[N];

struct Node{
    int l,r,cnt;
}tr[N * 50];

void add(int x,int y){
    ne[num] = h[x],e[num] = y,h[x] = num++;
}

int LCA(int a,int b){
    if(dep[a] < dep[b]) swap(a,b);
    for(int i = 19;i >= 0;i--)
        if(dep[f[a][i]] >= dep[b])
            a = f[a][i];
    if(a == b) return a;
    for(int i = 19;i >= 0;i--)
        if(f[a][i] != f[b][i])
            a = f[a][i],b = f[b][i];
    return f[a][0];
}


int build(int l,int r){
    int p = ++idx;
    if(l == r) return p;
    int mid = l + r >> 1;
    tr[p].l = build(l,mid);
    tr[p].r = build(mid + 1,r);
    return p;
}

int insert(int p,int l,int r,int x){
    int q = ++idx;
    tr[q] = tr[p];
    if(l == r){
        tr[q].cnt++;
        return q;
    }
    int mid = l + r >> 1;
    if(x <= mid) tr[q].l = insert(tr[q].l,l,mid,x);
    else tr[q].r = insert(tr[q].r,mid + 1,r,x);
    tr[q].cnt = tr[tr[q].l].cnt + tr[tr[q].r].cnt;
    return q;
}

void dfs(int u,int fa){
    if(id[u].size()){
        root[u] = insert(root[fa],1,m,id[u][0]);
        for(int i = 1;i < id[u].size();i++)
            root[u] = insert(root[u],1,m,id[u][i]);
    }
    else root[u] = root[fa];
    f[u][0] = fa,dep[u] = dep[fa] + 1;
    for(int i = 1;i < 20;i++)
        f[u][i] = f[f[u][i - 1]][i - 1];
    for(int i = h[u];~i;i = ne[i]) if(e[i] != fa) dfs(e[i],u);
}

int query(int p,int q,int a,int b,int l,int r,int k){
    if(l == r) return r;
    int cnt = tr[tr[p].l].cnt + tr[tr[q].l].cnt - tr[tr[a].l].cnt - tr[tr[b].l].cnt;
    int mid = l + r >> 1;
    if(k <= cnt) return query(tr[p].l,tr[q].l,tr[a].l,tr[b].l,l,mid,k);
    else return query(tr[p].r,tr[q].r,tr[a].r,tr[b].r,mid + 1,r,k - cnt);
}

int main() {
    read(n),read(m),read(q);
    memset(h,-1,sizeof h);
    for(int i = 1;i < n;i++){
        int x,y;
        read(x),read(y);
        add(x,y),add(y,x);
    }
    for(int i = 1;i <= m;i++) {
        int x;
        read(x);
        id[x].pb(i);
    }
    root[0] = build(1,m);
    dfs(1,0);   
    while(q--){
        int u,v,a;
        read(u),read(v),read(a);
        int lca = LCA(u,v);
        VI ans;
        int res = tr[root[u]].cnt + tr[root[v]].cnt -  tr[root[lca]].cnt - tr[root[f[lca][0]]].cnt;
        for(int i = 1;i <= min(res,a);i++){
            int x = query(root[u],root[v],root[lca],root[f[lca][0]],1,m,i);
            ans.pb(x);
        }
        printf("%d ",(int)ans.size());
        for(int x:ans) printf("%d ",x);
        puts("");
    }
    return 0;
}