【LG5055】可持久化文艺平衡树

【LG5055】可持久化文艺平衡树

题面

洛谷

题解

终于不可以用\(Trie\)水了。。。

和普通的\(FHQ\;treap\)差不多

注意一下\(pushdown\)、\(split\)要新开节点

代码

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <ctime>
using namespace std;
inline int gi() {
    register int data = 0, w = 1;
    register char ch = 0;
    while (!isdigit(ch) && ch != '-') ch = getchar();
    if (ch == '-') w = -1, ch = getchar();
    while (isdigit(ch)) data = 10 * data + ch - '0', ch = getchar();
    return w * data;
}
const int MAX_N = 2e5 + 5;
struct Node {
    int ls, rs, pri, val, size;
    long long sum;
    bool rev;
} t[MAX_N << 6];
int rub[MAX_N << 6], cur, rubtop, rt[MAX_N];
inline int newNode(int v = 0) {
    int o = rubtop ? rub[rubtop--] : ++cur;
    t[o].val = t[o].sum = v; t[o].pri = rand(); t[o].size = 1;
    t[o].ls = t[o].rs = t[o].rev = 0;
    return o;
}
inline int clone(int y) {
    int x = rubtop ? rub[rubtop--] : ++cur;
    t[x] = t[y];
    return x;
}
inline void pushup(int o) {
    t[o].size = t[t[o].ls].size + t[t[o].rs].size + 1;
    t[o].sum = t[t[o].ls].sum + t[t[o].rs].sum + t[o].val;
}
inline void pushdown(int o) {
    if (!t[o].rev) return ;
    swap(t[o].ls, t[o].rs);
    if (t[o].ls) t[o].ls = clone(t[o].ls), t[t[o].ls].rev ^= 1;
    if (t[o].rs) t[o].rs = clone(t[o].rs), t[t[o].rs].rev ^= 1;
    t[o].rev = 0;
}
void split(int o, int k, int &ls, int &rs) {
    if (!o) ls = rs = 0;
    else {
        pushdown(o);
        if (k <= t[t[o].ls].size) rs = clone(o), split(t[rs].ls, k, ls, t[rs].ls), pushup(rs);
        else ls = clone(o), split(t[ls].rs, k - t[t[o].ls].size - 1, t[ls].rs, rs), pushup(ls);
    }
}
int merge(int x, int y) {
    if (!(x && y)) return x ^ y;
    if (t[x].pri < t[y].pri) {
        pushdown(y);
        t[y].ls = merge(x, t[y].ls);
        pushup(y);
        return y;
    } else {
        pushdown(x);
        t[x].rs = merge(t[x].rs, y);
        pushup(x);
        return x;
    }
}
void insert(int &o, int k, int v) {
    int x, y;
    split(o, k, x, y);
    o = merge(merge(x, newNode(v)), y);
}
void erase(int &o, int pos) {
    int x, y, z;
    split(o, pos, x, z);
    split(x, pos - 1, x, y);
    rub[++rubtop] = y;
    o = merge(x, z);
}
void reverse(int &o, int l, int r) {
    int x, y, z;
    split(o, r, x, z);
    split(x, l - 1, x, y);
    t[y].rev ^= 1;
    o = merge(merge(x, y), z);
}
long long query(int &o, int l, int r) {
    int x, y, z;
    split(o, r, x, z);
    split(x, l - 1, x, y);
    long long res = t[y].sum;
    o = merge(merge(x, y), z);
    return res;
}
long long ans = 0;
int N;
int main () {
    srand(19260817);
    N = gi();
    for (int i = 1; i <= N; i++) {
        int v = gi(), op = gi();
        rt[i] = rt[v];
        if (op == 1) { int k = gi() ^ ans, val = gi() ^ ans; insert(rt[i], k, val); }
        else if (op == 2) { int k = gi() ^ ans; erase(rt[i], k); }
        else if (op == 3) { int l = gi() ^ ans, r = gi() ^ ans; reverse(rt[i], l, r); }
        else { int l = gi() ^ ans, r = gi() ^ ans; printf("%lld\n", ans = query(rt[i], l, r)); }
    }
    return 0;
}