可持久化数组用rope搞就MLE了有办法吗这个）

#include <bits/stdc++.h>

typedef std::pair <int, int> pii;
#define rep(i, x, y) for (int i = (x); i <= (y); ++i)
#define per(i, x, y) for (int i = (x); i >= (y); --i)

int read () {
    int res = 0;
    bool f = false;
    char temp = getchar();
    for (; !isdigit(temp); temp = getchar()) f = temp == '-';
    for (; isdigit(temp); temp = getchar()) res = res * 10 + temp - '0';
    if (f) return -res;
    return res;
}

#include <ext/rope>

constexpr int maxn = 2e5 + 5;
using namespace __gnu_cxx;
typedef rope <int> rpi;

int ts[maxn], n, m;
rpi *s[maxn], *sz[maxn];

int fd (rpi *r, int x) {
    int t = r->at(x);
    if (t != x) return fd(r, t);
    return x;
}

void uni (int u, int x, int y) {
    x = fd(s[u], x), y = fd(s[u], y);
    if (x == y) return;
    int sx = sz[u]->at(x), sy = sz[u]->at(y);
    if (sx > sy) s[u]->replace(y, x), sz[u]->replace(x, sx + sy);
    else s[u]->replace(x, y), sz[u]->replace(y, sx + sy);
}

signed main () {
    n = read(), m = read();
    rep (i, 0, n) ts[i] = i;
    s[0] = new rpi;
    s[0]->append(ts, ts + 1 + n);
    sz[0] = new rpi;
    sz[0]->append(n + 1, 1);
    rep (i, 1, m) {
        int op = read();
        if (op == 1) {
            int a = read(), b = read();
            s[i] = new rpi(*s[i - 1]);
            sz[i] = new rpi(*sz[i - 1]);
            uni(i, a, b);
        } else if (op == 2) {
            int a = read();
            s[i] = new rpi(*s[a]);
            sz[i] = new rpi(*sz[a]);
        } else {
            int a = read(), b = read();
            s[i] = new rpi(*s[i - 1]);
            sz[i] = new rpi(*sz[i - 1]);
            printf("%d\n", fd(s[i], a) == fd(s[i], b));
        }
    }
    return 0;
}