【洛谷 P2633】 Count on a tree（主席树，树上差分）

题目链接
思维难度0
实现难度7
建出主席树后用两点的状态减去lca和lca父亲的状态，然后在新树上跑第(k)小

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int MAXN = 100010;
const int MAXM = 100010;
inline int read(){
    int s = 0, w = 1;
    char ch = getchar();
    while(ch < '0' || ch > '9'){ if(ch == '-') w = -1; ch = getchar(); }
    while(ch >= '0' && ch <= '9'){ s = s * 10 + ch - '0'; ch = getchar(); }
    return s * w;
}
int cnt;
struct Pt{
    int lc, rc, val;
}t[MAXN * 21];
int build(int l, int r){
    int id = ++cnt;
    if(l == r) return id;
    int mid = (l + r) >> 1;
    t[id].lc = build(l, mid);
    t[id].rc = build(mid + 1, r);
    return id;
}
int update(int p, int l, int r, int x){
    int id = ++cnt; t[id] = t[p];
    if(l == r){ t[id].val++; return id; }
    int mid = (l + r) >> 1;
    if(x <= mid) t[id].lc = update(t[p].lc, l, mid, x);
    else t[id].rc = update(t[p].rc, mid + 1, r, x);
    t[id].val = t[t[id].lc].val + t[t[id].rc].val;
    return id;
}
struct Edge{
    int next, to;
}e[MAXN << 1];
int head[MAXN], num, w[MAXN], dep[MAXN], root[MAXN];
inline void Add(int from, int to){
    e[++num].to = to; e[num].next = head[from]; head[from] = num;
    e[++num].to = from; e[num].next = head[to]; head[to] = num;
}
int n, m, tot, f[MAXN][20];
struct lsh{
    int val, id;
    int operator < (const lsh A) const{
        return val < A. val;
    }
}p[MAXN];
void dfs(int u, int fa){
    root[u] = update(root[fa], 1, tot, w[u]);
    f[u][0] = fa; dep[u] = dep[fa] + 1;
    for(int i = head[u]; i; i = e[i].next)
        if(e[i].to != fa)
            dfs(e[i].to, u);
}
int a, b, c, val[MAXN], ans, lca;
int LCA(int u, int v){
    if(dep[u] > dep[v]) swap(u, v);
    int tmp = dep[v] - dep[u];
    for(int i = 0; i <= 19; ++i)
        if(tmp & (1 << i))
            v = f[v][i];
    for(int i = 19; ~i; --i)
        if(f[u][i] != f[v][i])
            u = f[u][i], v = f[v][i];
    return u == v ? u : f[u][0];
}
int solve(int l, int r, int a, int b, int c, int d, int k){
    if(l == r) return l;
    int lcnt = t[t[a].lc].val + t[t[b].lc].val - t[t[c].lc].val - t[t[d].lc].val, mid = (l + r) >> 1;
    if(lcnt < k) return solve(mid + 1, r, t[a].rc, t[b].rc, t[c].rc, t[d].rc, k - lcnt);
    else return solve(l, mid, t[a].lc, t[b].lc, t[c].lc, t[d].lc, k);
}
int main(){
    n = read(); m = read();
    for(int i = 1; i <= n; ++i)
        p[i].val = read(), p[i].id = i;
    sort(p + 1, p + n + 1);
    for(int i = 1; i <= n; ++i)
        if(p[i].val != p[i - 1].val){
            w[p[i].id] = ++tot;
            val[tot] = p[i].val;
        }
        else w[p[i].id] = tot;
    for(int i = 1; i < n; ++i)
        Add(read(), read());
    root[0] = build(1, tot); dfs(1, 0);
    for(int j = 1; j <= 19; ++j)
        for(int i = 1; i <= n; ++i)
            f[i][j] = f[f[i][j - 1]][j - 1];
    for(int i = 1; i <= m; ++i){
        a = read() ^ ans; b = read(); c = read(); lca = LCA(a, b);
        printf("%d
", ans = val[solve(1, tot, root[a], root[b], root[lca], root[f[lca][0]], c)]);
    }
    return 0;
}