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1、题目大意:就是给一个动态的森林求size域
2、分析: 这个就是一个动态树问题,对于每一个位置i有i+ki这个父亲, 于是这就是一个森林了,然后对于每一个修改直接lct维护就好,询问就是i到最外面的点的个数,恰好是一个链,就求一个 size就好(边界卡死啦啊啊啊
3、代码:

#include <stack> 
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
using namespace std;
#define LL long long
#define eps 1e-7
int n, m;
    struct Node{
        Node *ch[2], *fa;
        bool rev;
        int size;
        inline int which();
        inline bool reverse(){
            if(this) rev ^= 1;
        }
        inline void pd(){
            if(rev){
                swap(ch[0], ch[1]);
                ch[0] -> reverse();
                ch[1] -> reverse();
                rev = false;
            }
        } 
        inline void maintain(){
            size = 1;
            size += ch[0] -> size;
            size += ch[1] -> size;
        } 
        Node();
    } *null = new Node, pos[200010];
    Node::Node(){
            ch[0] = ch[1] = null;
            fa = null;
            size = 1;
            rev = false;
    }
    inline int Node::which(){
        if(fa == null || (fa -> ch[0] != this && fa -> ch[1] != this)) return -1;
        return fa -> ch[1] == this;
    }
    inline void rotate(Node *o){
        Node *p = o -> fa;
        int l = o -> which(), r = l ^ 1;
        o -> fa = p -> fa;
        if(p -> which() != -1) p -> fa -> ch[p -> which()] = o;
        p -> ch[l] = o -> ch[r];
        if(o -> ch[r]) o -> ch[r] -> fa = p;
        o -> ch[r] = p; p -> fa = o;
        o -> ch[r] -> maintain();
        o -> maintain(); 
    } 
     
    inline void splay(Node* o){
        static stack<Node*> st;
        if(!o) return;
        Node* p = o;
        while(1){
            st.push(p);
            if(p -> which() == -1) break;
            p = p -> fa;
        }
        while(!st.empty()){
            st.top() -> pd(); st.pop();
        }
         
        while(o -> which() != -1){
            p = o -> fa;
            if(p -> which() != -1){
                if(p -> which() ^ o -> which()) rotate(o);
                else rotate(p);
            }
            rotate(o);
        }
    }
     
    inline void Access(Node* o){
        Node* p = null;
        while(o != null){
            splay(o);
            o -> ch[1] = p;
            o -> maintain();
            p = o; o = o -> fa;
        }
    }
     
    inline void Link(Node* x, Node* y){
        Access(x);
        splay(x);
        if(x -> ch[0]) x -> ch[0] -> fa = null; 
        x -> ch[0] = null;
        x -> fa = y;
        x -> maintain();
    }
 
int main(){
    null->ch[0]=null->ch[1]=null->fa=NULL;
    null->size=0;
    scanf("%d", &n);
    for(int i = 1; i <= n; i ++){
        int y; scanf("%d", &y);
        if(i + y <= n) pos[i].fa = &pos[i + y];
    }
    scanf("%d", &m);
    while(m --){
        int op, x, y;
        scanf("%d", &op);
        if(op == 1){
            scanf("%d", &x); x ++;
            Access(&pos[x]);
            splay(&pos[x]);
            printf("%d\n", pos[x].size);
        }
        else{
            scanf("%d%d", &x, &y); x ++;
            if(x + y <= n)   Link(&pos[x], &pos[x+y]);
            else Link(&pos[x], null);
        }
    }
    return 0;
}