Solution

要離線來做。
考慮最後一遍是按樹的編號順序一遍掃過。
所以就是要先把兩棵樹之間要改變的資訊都維護出來。
這道題是對生長節點建出一個虛點。
就可以在多棵樹上同時連邊。

#include <bits/stdc++.h>
using namespace std;

const int N = 404040;

inline char get(void) {
    static char buf[100000], *S = buf, *T = buf;
    if (S == T) {
        T = (S = buf) + fread(buf, 1, 100000, stdin);
        if
 
 (S == T) return EOF;
    }
    return *S++;
}
inline void read(int &x) {
    static char c; x = 0;
    for (c = get(); c < '0' || c > '9'; c = get());
    for (; c >= '0' && c <= '9'; c = get()) x = x * 10 + c - '0';
}

struct node *null;
struct node {
    node *ch[2];
    node *fa;
    int
 
 size, key, rev;
    inline void Reverse(void) {
        rev ^= 1; swap(ch[1], ch[0]);
    }
    inline void PushUp(void) {
        size = key + ch[0]->size + ch[1]->size;
    }
    inline void PushDown(void) {
        if (rev) {
            ch[0]->Reverse();
            ch[1]->Reverse();
            rev = 0
 
;
        }
    }
    inline void New(int k) {
        fa = ch[0] = ch[1] = null;
        key = k; rev = 0; PushUp();
    }
};
node *p;
node mem[1];
node T[N];
int n, m, opt, x, y, z, c1, cnt, lst, Ocnt;
int L[N], R[N], to[N];
int ans[N];
struct Option {
    int pos, id, x, y;
    Option(int p = 0, int i = 0, int _x = 0, int _y = 0):pos(p), id(i), x(_x), y(_y) {}
    inline friend bool operator <(const Option &a, const Option &b) {
        return a.pos == b.pos ? a.id < b.id : a.pos < b.pos;
    }
};
Option Q[N];

inline bool IsRoot(node *x) {
    return x->fa == null || (x->fa->ch[0] != x && x->fa->ch[1] != x);
}
inline void Rotate(node *x) {
    node *y = x->fa, *z = y->fa;
    int l = (y->ch[0] != x), r = l ^ 1;
    if (!IsRoot(y)) {
        if (z->ch[0] == y) z->ch[0] = x;
        else z->ch[1] = x;
    }
    x->fa = z; y->fa = x; x->ch[r]->fa = y;
    y->ch[l] = x->ch[r]; x->ch[r] = y;
    y->PushUp(); x->PushUp();
}
inline void Down(node *x) {
    if (!IsRoot(x)) Down(x->fa);
    x->PushDown();
}
inline void Splay(node *x) {
    Down(x);
    while (!IsRoot(x)) {
        node *y = x->fa, *z= y->fa;
        if (!IsRoot(y)) {
            if (z->ch[0] == y ^ y->ch[0] == x) Rotate(x);
            else Rotate(y);
        }
        Rotate(x);
    }
}
inline node *Access(node *x) {
    node *y;
    for (y = null; x != null; x = x->fa) {
        Splay(x); x->ch[1] = y;
        x->PushUp(); y = x;
    }
    return y;
}
inline void MakeRoot(node *x) {
    Access(x); Splay(x); x->Reverse();
}
inline void Link(node *x, node *y) {
    MakeRoot(x); x->fa = y;
}
inline void Cut(node *x, node *y) {
    MakeRoot(x); Access(y); Splay(y);
    x->fa = y->ch[0] = null; y->PushUp();
}
inline void Cut(node *x) {
    Access(x); Splay(x);
    x->ch[0] = x->ch[0]->fa = null;
    x->PushUp();
}
void Init(void) {
    null = mem;
    null->fa = null->ch[0] = null->ch[1] = null;
    null->size = null->key = 0;
}

int main(void) {
    read(n); read(m); c1 = 1; Init();
    to[c1] = L[c1] = 1; R[c1] = n;
    T[++cnt].New(1);
    T[lst = ++cnt].New(0);
    Link(T + 2, T + 1);
    for (int i = 1; i <= m; i++) {
        read(opt);
        if (opt == 0) {
            read(L[++c1]); read(R[c1]);
            T[++cnt].New(1); to[c1] = cnt;
            Q[++Ocnt] = Option(1, i - m, cnt, lst);
        } else if (opt == 1) {
            read(x); read(y); read(z);
            x = max(x, L[z]); y = min(y, R[z]);
            if (x <= y) {
                T[++cnt].New(0);
                if (x > 1) Link(T + cnt, T + lst);
                Q[++Ocnt] = Option(x, i - m, cnt, to[z]);
                Q[++Ocnt] = Option(y + 1, i - m, cnt, lst);
                lst = cnt;
            }
        } else {
            read(x); read(y); read(z);
            Q[++Ocnt] = Option(x, i, to[y], to[z]);
        }
    }
    for (int i = 1; i <= m; i++) ans[i] = -1;
    sort(Q + 1, Q + Ocnt + 1);
    for (int i = 1, p1 = 1; i <= m; i++)
        for (; p1 <= Ocnt && Q[p1].pos == i; p1++) {
            if (Q[p1].id > 0) {
                Access(T + Q[p1].x); Splay(T + Q[p1].x); ans[Q[p1].id] = T[Q[p1].x].size;
                p = Access(T + Q[p1].y); Splay(T + Q[p1].y); ans[Q[p1].id] += T[Q[p1].y].size;
                Access(p); Splay(p); ans[Q[p1].id] -= 2 * p->size;
            } else {
                Cut(T + Q[p1].x); Link(T + Q[p1].x, T + Q[p1].y);
            }
        }
    for (int i = 1; i <= m; i++)
        if (~ans[i]) printf("%d\n", ans[i]);
    return 0;
}