【可持久化线段树】[SPOJ COT]Count on a tree

题目大意：给定一棵树，然后询问连个节点间路径上的权值的第K小的权值大小
题目分析：和普通的第K大的可持久化线段树差距不大，但是要写个LCA可以发现Tree(a)+Tree(b)−Tree(LCA)−LCA(fa[LCA])就是两个节点之间的线段树了，然后按照普通的可持久化线段树搞一搞就好了

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int MAXN = 1e5+100;
struct node{
    int ch[2];
    int sum;
}pool[MAXN*21];
int roots[MAXN+10], q[MAXN+10], p[MAXN+10], tot, nlen;
struct Ed{
    int u, v;
    Ed *next;
}Edges[MAXN*2+10], *ecnt=Edges, *adj[MAXN+10];
int Fa[MAXN+10][20], Dep[MAXN+10];
void addedge(int u, int v){
    ++ecnt;
    ecnt->v = v;
    ecnt->next = adj[u];
    adj[u] = ecnt;
}
void Insert(int &u, int l, int r, int v){
    ++tot;
    pool[tot] = pool[u];
    u = tot;
    pool[u].sum ++;
    if(l == r) return ;
    int mid = (l + r) >> 1;
    if(v <= mid) Insert(pool[u].ch[0], l, mid, v);
    else Insert(pool[u].ch[1], mid+1, r, v);
}
void dfs(int u, int fa){
    Fa[u][0] = fa;
    for(int i=0;i<18;i++)
        Fa[u][i+1] = Fa[Fa[u][i]][i];
    Dep[u] = Dep[fa] + 1;
    roots[u] = roots[fa];
    Insert(roots[u], 1, nlen, q[u]);
    for(Ed *p=adj[u];p;p=p->next){
        if(p->v == fa) continue;
        dfs(p->v, u);
    }
}
int LCA(int a, int b){
    if(Dep[a] > Dep[b]) swap(a, b);
    for(int k=18;k>=0;k--)
        if(Dep[a] <= Dep[Fa[b][k]])
            b = Fa[b][k];
    if(a == b) return a;
    for(int k=18;k>=0;k--){
        if(Fa[a][k] != Fa[b][k] && Fa[a][k] != -1){
            a = Fa[a][k];
            b = Fa[b][k];
        }
    }
    return Fa[a][0];
}
int Query(int lroot, int rroot, int lcroot, int lcfroot, int l, int r, int k){
    if(l >= r) return l;
    int mid = (l + r) >> 1;
    int cz = pool[pool[lroot].ch[0]].sum;
    cz += pool[pool[rroot].ch[0]].sum;
    cz -= pool[pool[lcroot].ch[0]].sum;
    cz -= pool[pool[lcfroot].ch[0]].sum;
    if(k <= cz)
        return Query(pool[lroot].ch[0], pool[rroot].ch[0], pool[lcroot].ch[0], pool[lcfroot].ch[0], l, mid, k);
    return Query(pool[lroot].ch[1], pool[rroot].ch[1], pool[lcroot].ch[1], pool[lcfroot].ch[1], mid+1, r, k-cz);
}
int main(){
    memset(Fa,-1, sizeof Fa);
    int n, m, a, b, k;
    scanf("%d%d", &n, &m);
    for(int i=1;i<=n;i++){
        scanf("%d", &q[i]);
        p[i] = q[i];
    }
    sort(p+1, p+1+n);
    nlen = unique(p+1, p+1+n) - (p+1);
    for(int i=1;i<=n;i++)
        q[i] = lower_bound(p+1, p+1+nlen, q[i]) - p;
    for(int i=2;i<=n;i++){
        scanf("%d%d", &a, &b);
        addedge(a, b);
        addedge(b, a);
    }
    dfs(1, 0);
    for(int i=1;i<=m;i++){
        scanf("%d%d%d", &a, &b, &k);
        int lc = LCA(a, b);
        printf("%d\n", p[Query(roots[a], roots[b], roots[lc], roots[Fa[lc][0]], 1, nlen, k)]);
    }

    return 0;
}
/*
2 1
1 2
1 2
1 2 2
*/