bzoj 2002 LCT

LCT最基础的题，就用到了一个ACCESS操作

首先我们将这个绵羊弹飞的情况看成一颗树，那么假设X点被弹飞到

Y点，那么Y为X的父亲节点，弹飞的话父亲节点为n+1（虚设）

那么每个询问就是询问X点到根节点n+1的路径长度（节点数）

每个修改操作就是将以X为根节点的子树和X的父亲断开，连接到Y上

这样简单的维护森林连通性的问题，动态树中的LCT解决就行了

/**************************************************************
    Problem:
    User: BLADEVIL
    Language: Pascal
    Result: Accepted
    Time: ms
    Memory: kb
****************************************************************/

//By BLADEVIL
var
    n, m                :longint;
    father, size        :array[-..] of longint;
    son                 :array[-..,..] of longint;
    root                :array[-..] of boolean;

procedure update(x:longint);
begin
    size[x]:=size[son[x,]]+size[son[x,]]+;
end;

procedure left_rotate(x:longint);
var
    y                   :longint;
begin
    y:=son[x,];
    son[x,]:=son[y,];
    father[son[x,]]:=x;
    son[y,]:=x;
    if x=son[father[x],] then
        son[father[x],]:=y else
    if x=son[father[x],] then
        son[father[x],]:=y;
    father[y]:=father[x];
    father[x]:=y;
    root[y]:=root[x] xor root[y];
    root[x]:=root[x] xor root[y];
    update(x); update(y);
end;

procedure right_rotate(x:longint);
var
    y                   :longint;
begin
    y:=son[x,];
    son[x,]:=son[y,];
    father[son[x,]]:=x;
    son[y,]:=x;
    if x=son[father[x],] then
        son[father[x],]:=y else
    if x=son[father[x],] then
        son[father[x],]:=y;
    father[y]:=father[x];
    father[x]:=y;
    root[y]:=root[y] xor root[x];
    root[x]:=root[y] xor root[x];
    update(x); update(y);
end;

procedure splay(x:longint);
begin
    while not root[x] do
        if x=son[father[x],] then
            left_rotate(father[x]) else
            right_rotate(father[x]);
end;

procedure access(x:longint);
var
    y                   :longint;
begin
    splay(x);
    while father[x]<> do
    begin
        y:=father[x];
        splay(y);
        root[son[y,]]:=true;
        root[x]:=false;
        son[y,]:=x;
        update(y);
        splay(x);
    end;
end;

procedure init;
var
    i                   :longint;
begin
    read(n);
    for i:= to n do
    begin
        read(father[i]);
        father[i]:=father[i]+i;
        if father[i]>n then father[i]:=n+;
    end;
    read(m);
end;

procedure main;
var
    i                   :longint;
    x, y, z             :longint;
begin
    for i:= to n+ do size[i]:=;
    fillchar(root,sizeof(root),true);
    for i:= to m do
    begin
        read(x);
        if x= then
        begin
            read(y); inc(y);
            access(y);
            writeln(size[son[y,]]);
        end else
        begin
            read(y,z); inc(y);
            splay(y);
            father[son[y,]]:=father[y];
            root[son[y,]]:=true;
            son[y,]:=;
            size[y]:=size[son[y,]]+;
            father[y]:=y+z;
            if father[y]>n then father[y]:=n+;
        end;
    end;
end;

begin
    init;
    main;
end.