[SPOJ10628]Count on a tree（主席树）

Posted 2021-02-28 zcr-blog

题目链接

题目描述

给你一棵有n个结点的树，节点编号为1~n。

每个节点都有一个权值。

要求执行以下操作：

U V K：求从节点u到节点v的第k小权值。

Solution

树上主席树裸题。

思路和序列差不多，树上前缀和即可。

可持久化时的前一个版本就是它的父亲。

设查询(u,v)，值就是u+v-lca-fa[lca]

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
const int N = 100010;
struct chairman_tree{
    int val;
    int ls, rs;
}tr[N * 20];
int rt[N], tot;
struct node{
    int pre, to;
}edge[N << 1];
int head[N], tt;
int n, m, a[N], b[N], nn;
int sz[N], bl[N], dep[N], f[N], son[N], tn[N];
int lastans;
void ins(int &cur, int pre, int l, int r, int pos) {
    cur = ++tot;
    tr[cur] = tr[pre];
    tr[cur].val++;
    if (l == r) return;
    int mid = (l + r) >> 1;
    if (pos <= mid) ins(tr[cur].ls, tr[pre].ls, l, mid, pos);
    else ins(tr[cur].rs, tr[pre].rs, mid + 1, r, pos); 
}
int ask(int o, int p, int q, int s, int l, int r, int k) {
    if (l == r) return l;
    int mid = (l + r) >> 1;
    int num = tr[tr[o].ls].val + tr[tr[p].ls].val - tr[tr[q].ls].val - tr[tr[s].ls].val;
    if (k <= num) return ask(tr[o].ls, tr[p].ls, tr[q].ls, tr[s].ls, l, mid, k);
    else return ask(tr[o].rs, tr[p].rs, tr[q].rs, tr[s].rs, mid + 1, r, k - num);
}
void dfs(int x, int fa) {
    sz[x] = 1;
    son[x] = 0;
    ins(rt[x], rt[fa], 1, nn, a[x]);
    for (int i = head[x]; i; i = edge[i].pre) {
        int y = edge[i].to;
        if (y == fa) continue;
        dep[y] = dep[x] + 1;
        f[y] = x;
        dfs(y, x);
        sz[x] += sz[y];
        if (sz[y] > sz[son[x]]) {
            son[x] = y;
        }
    }
}
void dfs2(int x, int chain) {
    bl[x] = chain;
    if (son[x]) dfs2(son[x], chain);
    for (int i = head[x]; i; i = edge[i].pre) {
        int y = edge[i].to;
        if (y != f[x] && y != son[x]) {
            dfs2(y, y);
        }
    }
}
int LCA(int u, int v) {
    while (bl[u] != bl[v]) {
        if (dep[bl[u]] < dep[bl[v]]) swap(u, v);
        u = f[bl[u]];
    }
    return dep[u] < dep[v] ? u : v;
}
void add(int u, int v) {
    edge[++tt] = node{head[u], v};
    head[u] = tt; 
}
int main() {
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++) {
        scanf("%d", &a[i]);
        b[++nn] = a[i];
    }
    sort(b + 1, b + nn + 1);
    nn = unique(b + 1, b + nn + 1) - b - 1;
    for (int i = 1; i <= n; i++) {
        int p = lower_bound(b + 1, b + nn + 1, a[i]) - b;
        tn[p] = a[i];
        a[i] = p;
    }
    for (int i = 1, u, v; i < n; i++) {
        scanf("%d%d", &u, &v);
        add(u, v);
        add(v, u);
    }
    dfs(1, 0);
    dfs2(1, 1);
    while (m--) {
        int u, v, k;
        scanf("%d%d%d", &u, &v, &k);
        u ^= lastans;
        int lca = LCA(u, v);
        lastans = tn[ask(rt[u], rt[v], rt[lca], rt[f[lca]], 1, nn, k)];
        printf("%d
", lastans);
    }
    return 0;
}