51nod1648 洞 LCT

非常简单的一眼LCT，然而我没有在20min内码完，太失败了...

第一问，直接查根的前驱

第二问，查链的子树大小

复杂度$O((n + m) log n)$

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

extern inline char gc() {
    static char RR[], *S = RR + , *T = RR + ;
    if(S == T) fread(RR, , , stdin), S = RR;
    return *S ++;
}
inline int read() {
    int p = , w = ; char c = gc();
    while(c > '' || c < '') { if(c == '-') w = -; c = gc(); }
    while(c >= '' && c <= '') p = p *  + c - '', c = gc();
    return p * w;
}

int wr[], rw;
#define pc(o) *W ++ = (o)
char WR[], *W = WR;
inline void write(int x, char c) {
    if(!x) pc('');
    if(x < ) x = -x, pc('-');
    while(x) wr[++ rw] = x % , x /= ;
    while(rw) pc(wr[rw --] + ''); pc(c);
}

#define ri register int
#define sid 200500

#define ls(o) s[o][0]
#define rs(o) s[o][1]
int n, m, a[sid];
int fa[sid], s[sid][], rev[sid], sz[sid]; 

inline bool isr(int o) { return ls(fa[o]) != o && rs(fa[o]) != o; }
inline bool isrc(int o) { return rs(fa[o]) == o; }
inline void upd(int o) { sz[o] = sz[ls(o)] + sz[rs(o)] + ; }

inline void rotate(int o) {
    int f = fa[o], g = fa[f];
    int ro = isrc(o), rf = isrc(f);
    fa[o] = g; if(!isr(f)) s[g][rf] = o;
    if(s[o][ro ^ ]) fa[s[o][ro ^ ]] = f;
    s[f][ro] = s[o][ro ^ ]; fa[f] = o; s[o][ro ^ ] = f;
    upd(f); upd(o);
}

inline void prev(int o) {
    swap(ls(o), rs(o));
    rev[o] ^= ;
}

inline void prv(int o) {
    if(!rev[o]) return;
    prev(ls(o)); prev(rs(o)); rev[o] = ;
}

inline void pushrev(int o) {
    if(!isr(o)) pushrev(fa[o]);
    prv(o);
}

void splay(int o) {
    pushrev(o);
    while(!isr(o)) {
        int f = fa[o];
        if(!isr(f)) rotate(isrc(o) == isrc(f) ? f : o);
        rotate(o);
    }
}

void access(int o) {
    for(ri t = ; o; t = o, o = fa[o])
    splay(o), s[o][] = t, upd(o);
}

inline void makeroot(int o) {
    access(o); splay(o); prev(o);
}

inline void link(int u, int v) {
    makeroot(u); fa[u] = v;
}

inline void cut(int u, int v) {
    makeroot(u); access(v); splay(v);
    s[v][] = ; fa[u] = ; upd(v);
}

inline int get(int u) {
    prv(u); u = s[u][]; prv(u);
    while(s[u][]) u = s[u][], prv(u);
    return u;
}

inline void qry(int u) {
    makeroot(n + ); access(u); splay(n + );
    int ans2 = sz[n + ] - , ans1 = get(n + );
    splay(ans1); write(ans1, ' '); write(ans2, '\n');
}

int main() {
    n = read(); m = read();
    for(ri i = ; i <= n; i ++) {
        a[i] = read();
        if(i + a[i] <= n) link(i, i + a[i]);
        else link(i, n + );
    }
    for(ri i = ; i <= m; i ++) {
        int opt = read(), x = read();
        if(opt == ) {
            if(x + a[x] <= n) cut(x, x + a[x]); else cut(x, n + ); int y = read();
            a[x] = y; if(x + a[x] <= n) link(x, x + a[x]); else link(x, n + );
        } else qry(x);
    }
    fwrite(WR, , W - WR, stdout);
    return ;
}