[HNOI2010]弹飞绵羊

嘟嘟嘟


挺好的LCT题。
所以我决定不拿分块水过，按正解写一发。


先新建一个虚拟节点\(n + 1\)，跳到这个点就说明被弹飞了。
1.建树：
\(i\)向\(min \{ i + k_i, n + 1 \}\)连边。
2.修改
断掉\(i\)和\(min \{ i + k_i, n + 1 \}\)的边，并链接\(i\)和\(min \{ i + k_{new}, n + 1 \}\)，然后把\(k_i\)赋成\(k_{new}\)。
3.查询
即树上\(i\)和\(n + 1\)的链长，别忘减\(1\)。

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
#define enter puts("") 
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define In inline
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 2e5 + 5;
inline ll read()
{
  ll ans = 0;
  char ch = getchar(), last = ' ';
  while(!isdigit(ch)) last = ch, ch = getchar();
  while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
  if(last == '-') ans = -ans;
  return ans;
}
inline void write(ll x)
{
  if(x < 0) x = -x, putchar('-');
  if(x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}

int n, m, a[maxn];
struct Tree
{
  int ch[2], fa;
  int siz, rev;
}t[maxn];

In void change(int now)
{
  swap(t[now].ch[0], t[now].ch[1]);
  t[now].rev ^= 1;
}
In void pushdown(int now)
{
  if(t[now].rev)
    {
      if(t[now].ch[0]) change(t[now].ch[0]);
      if(t[now].ch[1]) change(t[now].ch[1]);
      t[now].rev = 0;
    }
}
In void pushup(int now)
{
  t[now].siz = t[t[now].ch[0]].siz + t[t[now].ch[1]].siz + 1;
}
In bool n_root(int now)
{
  return t[t[now].fa].ch[0] == now || t[t[now].fa].ch[1] == now;
}
In void rotate(int x)
{
  int y = t[x].fa, z = t[y].fa, k = (t[y].ch[1] == x);
  if(n_root(y)) t[z].ch[t[z].ch[1] == y] = x; t[x].fa = z;
  t[y].ch[k] = t[x].ch[k ^ 1]; t[t[y].ch[k]].fa = y;
  t[x].ch[k ^ 1] = y; t[y].fa = x;
  pushup(y), pushup(x);
}
int st[maxn], top = 0;
In void splay(int x)
{
  int y = x;
  st[top = 1] = y;
  while(n_root(y)) y = t[y].fa, st[++top] = y;
  while(top) pushdown(st[top--]);
  while(n_root(x))
    {
      int y = t[x].fa, z = t[y].fa;
      if(n_root(y)) rotate(((t[z].ch[0] == y) ^ (t[y].ch[0] == x)) ? x : y);
      rotate(x);
    }
}
In void access(int x)
{
  int y = 0;
  while(x)
    {
      splay(x); t[x].ch[1] = y;
      pushup(x);
      y = x; x = t[x].fa;
    }
}
In void make_root(int x)
{
  access(x); splay(x);
  change(x);
}
In int find_root(int x)
{
  access(x); splay(x);
  while(t[x].ch[0]) pushdown(x), x = t[x].ch[0];
  return x;
}
In void Link(int x, int y)
{
  make_root(x);
  if(find_root(y) != x) t[x].fa = y;
}
In void Cut(int x, int y)
{
  make_root(x);
  if(find_root(y) == x && t[x].fa == y && !t[x].ch[1])
    t[x].fa = t[y].ch[0] = 0, pushup(y);
}
In int query(int x)
{
  make_root(x);
  access(n + 1); splay(n + 1);
  return t[n + 1].siz - 1;
}

int main()
{
  n = read();
  for(int i = 1; i <= n; ++i)
    {
      a[i] = read();
      Link(i, min(a[i] + i, n + 1));
    }
  m = read();
  for(int i = 1; i <= m; ++i)
    {
      int pos = read(), x = read() + 1;
      if(pos == 1) write(query(x)), enter;
      else
	{
	  int k = read();
	  Cut(x, min(x + a[x], n + 1));
	  Link(x, min(x + k, n + 1));
	  a[x] = k;
	}
    }
  return 0;
}