题链：
http://www.lydsy.com/JudgeOnline/problem.php?id=2002
题解：
LCT
如果把弹跳的起点和终点连一条边，弹出去的与n+1号点连边，
则不难发现，整个图形成了一颗树，
同时需要支持树的修改(拆分，合并)和询问点的深度(该点到根的链上的点的个数)，
所以LCT可以很轻松的解决本题。
代码：
#include<cstdio>
#include<cstring>
#include<iostream>
#include<cstdlib>
#define MAXN 200050
using namespace std;
int N,M,next[MAXN];
struct LCT{int ch[MAXN][2],fa[MAXN],rev[MAXN],size[MAXN];bool Which(int x){return ch[fa[x]][1]==x;}void Reverse(int x){swap(ch[x][0],ch[x][1]);rev[x]^=1;}bool Isroot(int x){return ch[fa[x]][0]!=x&&ch[fa[x]][1]!=x;}void Pushup(int x){size[x]=size[ch[x][0]]+size[ch[x][1]]+1;}void Pushdown(int x){if(!Isroot(x)) Pushdown(fa[x]);if(rev[x]) Reverse(ch[x][0]),Reverse(ch[x][1]),rev[x]^=1;}void Rotate(int x){static int y,z,l1,l2;y=fa[x]; z=fa[y];l1=Which(y); l2=Which(x); fa[x]=z;if(!Isroot(y)) ch[z][l1]=x;fa[y]=x; fa[ch[x][l2^1]]=y; ch[y][l2]=ch[x][l2^1]; ch[x][l2^1]=y;Pushup(y);}void Splay(int x){static int y; Pushdown(x);for(;y=fa[x],!Isroot(x);Rotate(x)) if(!Isroot(y)) Rotate(Which(y)==Which(x)?y:x);Pushup(x);}void Access(int x){static int y;for(y=0;x;y=x,x=fa[x]) Splay(x),ch[x][1]=y,Pushup(x);//!!!}void Beroot(int x){Access(x); Splay(x); Reverse(x);}void Cut(int x,int y){Beroot(x); Access(y); Splay(y);fa[x]=ch[y][0]=0; Pushup(y);}void Link(int x,int y){Beroot(x); fa[x]=y;}void Modify(int x,int k){static int t; t=min(x+k,N+1);Cut(next[x],x); Link(t,x); next[x]=t;}int Query(int x){Beroot(N+1); Access(x); Splay(x);return size[x]-1;}
}DT;
int main(){scanf("%d",&N); DT.size[N+1]=1;for(int i=1,k;i<=N;i++){scanf("%d",&k); next[i]=min(i+k,N+1);DT.fa[i]=next[i]; DT.size[i]=1;} scanf("%d",&M); int a,b,c; for(int i=1;i<=M;i++){scanf("%d%d",&a,&b); b++;if(a==1) printf("%d\n",DT.Query(b));else scanf("%d",&c),DT.Modify(b,c);}return 0;
}