【LN2014】LCA

题目链接：https://www.luogu.com.cn/problem/P4211

题目大意：给定一棵有根树，对于 \(q\) 个询问 \(l, r, z\), 求 \(\sum\limits_{l \leq i \leq r} {depth(Lca(i, z))}\)

solution

考虑另一个问题：求出\(\sum\limits_{1 \leq i \leq n} {dep[Lca(i, z)]}\)此问题可以转化为 \(i\) 从 1 到 \(n\), 分别把 1 到 \(i\) 路径上的所有点权值加 1, 再求 \(z\) 点的权值

不难发现 , 原问题可已转化为\(\sum\limits_{1 \leq i \leq r} {dep[Lca(i, z)]} - \sum\limits_{1 \leq i \leq l - 1} {dep[Lca(i, z)]}\) , 可以把原问题分解成两个子问题并离线 , 从 1 到 \(n\) 分别用树剖修改 , 在此过程中算出子问题的答案 , 最后统计即可

时间复杂度: \(O(nlog^2n)\)

code

#include<bits/stdc++.h>
using namespace std;
template <typename T> inline void read(T &FF) {
    int RR = 1; FF = 0; char CH = getchar();
    for(; !isdigit(CH); CH = getchar()) if(CH == '-') RR = -RR;
    for(; isdigit(CH); CH = getchar()) FF = FF * 10 + CH - 48;
    FF *= RR;
}
inline void file(string str) {
    freopen((str + ".in").c_str(), "r", stdin);
    freopen((str + ".out").c_str(), "w", stdout);
}
#define mod 201314
const int N = 1e5 + 10;
int now, fst[N], nxt[N], num[N], l[N], r[N], z[N], rs[N], dep[N];
int n, q, si, ni, pi = 1, size[N], son[N], fa[N], tp[N], rev[N], sg[N]; 
void add(int u, int v) {
    nxt[++now] = fst[u], fst[u] = now, num[now] = v;
    nxt[++now] = fst[v], fst[v] = now, num[now] = u;
}
struct Segment_tree{
    int val, tag;
}xds[N << 2];
struct Que{
    int qi, fi, id, fl;
    friend bool operator < (Que ai, Que bi) {
        return ai.qi < bi.qi;
    }
}qy[N];
void push_up(int pos) {
    xds[pos].val = (xds[pos << 1].val + xds[pos << 1 | 1].val) % mod;
}
void Add(int pos, int l, int r, int ki) {
    xds[pos].val = (xds[pos].val + (r - l + 1) * ki % mod) % mod;
    xds[pos].tag = (xds[pos].tag + ki) % mod;
}
void push_down(int pos, int l, int r) {
    if(xds[pos].tag == 0) return;
    int mid = (l + r) >> 1;
    Add(pos << 1, l, mid, xds[pos].tag);
    Add(pos << 1 | 1, mid + 1, r, xds[pos].tag);
    xds[pos].tag = 0;
}
void modify(int pos, int l, int r, int ll, int rr) {
    if(l >= ll && r <= rr) {
        Add(pos, l, r, 1);
        return;
    }
    push_down(pos, l, r); int mid = (l + r) >> 1;
    if(mid >= ll) modify(pos << 1, l, mid, ll, rr);
    if(mid < rr) modify(pos << 1 | 1, mid + 1, r, ll, rr);
    push_up(pos);
} 
int query(int pos, int l, int r, int ll, int rr) {
    if(l >= ll && r <= rr) return xds[pos].val;
    push_down(pos, l, r); int mid = (l + r) >> 1, res = 0;
    if(mid >= ll) res = (res + query(pos << 1, l, mid, ll, rr)) % mod;
    if(mid < rr) res = (res + query(pos << 1 | 1, mid + 1, r, ll, rr)) % mod;
//  push_up(pos);
    return res;
}
void tree_dsu(int xi) {
    size[xi] = 1; dep[xi] = dep[fa[xi]] + 1;
    for(int i = fst[xi]; i; i = nxt[i])
        if(num[i] != fa[xi]) {
            tree_dsu(num[i]);
            size[xi] += size[num[i]];
            if(size[num[i]] > size[son[xi]])
                son[xi] = num[i];
        }
}
void tree_pre(int xi) {
    if(son[xi]) {
        tp[son[xi]] = tp[xi], sg[++si] = son[xi];
        rev[son[xi]] = si, tree_pre(son[xi]);
    }
    for(int i = fst[xi]; i; i = nxt[i])
        if(num[i] != son[xi] && num[i] != fa[xi]) {
            tp[num[i]] = num[i], sg[++si] = num[i];
            rev[num[i]] = si, tree_pre(num[i]);
        }
}
void modify_list(int xi) {
    int yi = 1;
    while(tp[xi] != tp[yi]) {
        if(dep[tp[xi]] < dep[tp[yi]]) swap(xi, yi);
        modify(1, 1, n, rev[tp[xi]], rev[xi]);
        xi = fa[tp[xi]];
    }
    if(dep[xi] > dep[yi]) swap(xi, yi);
    modify(1, 1, n, rev[xi], rev[yi]);
}
int query_list(int xi) {
    int yi = 1, res = 0;
    while(tp[xi] != tp[yi]) {
        if(dep[tp[xi]] < dep[tp[yi]]) swap(xi, yi);
        res = (res + query(1, 1, n, rev[tp[xi]], rev[xi])) % mod;
        xi = fa[tp[xi]];
    }
    if(dep[xi] > dep[yi]) swap(xi, yi);
    res = (res + query(1, 1, n, rev[xi], rev[yi])) % mod;
    return res;
}
int main() {
    //file("");
    read(n), read(q);
    for(int i = 2; i <= n; i++) {
        read(fa[i]); fa[i]++;
        add(fa[i], i);
    }
    tp[1] = rev[1] = sg[1] = si = 1;
    tree_dsu(1), tree_pre(1);
    for(int i = 1; i <= q; i++) {
        read(l[i]), read(r[i]), read(z[i]); l[i]++, r[i]++, z[i]++;
        if(l[i] != 1) qy[++ni].qi = l[i] - 1, qy[ni].id = i, qy[ni].fi = z[i], qy[ni].fl = -1;
        qy[++ni].qi = r[i], qy[ni].id = i, qy[ni].fi = z[i], qy[ni].fl = 1;
    }
    sort(qy + 1, qy + ni + 1);
    for(int i = 1; i <= n; i++) {
        modify_list(i);
    //  cout << query(1, 1, n, sg[2], sg[2]) << endl;
        while(pi <= ni && qy[pi].qi == i)
            rs[qy[pi].id] += qy[pi].fl * query_list(qy[pi].fi), pi++;
    }
    for(int i = 1; i <= q; i++)
        cout << (rs[i] + mod) % mod << endl;
    return 0;
}

来源：https://www.cnblogs.com/magicduck/p/12238534.html