【BZOJ2049】洞穴勘测（LCT）

题意：一张图，要求支持以下操作：

1.加边

2.删边

3.询问两点之间是否联通

100%的数据满足n≤10000, m≤200000

思路：LCT裸题，不需要维护任何信息

 var t:array[..,..]of longint;
     fa,rev,q:array[..]of longint;
     n,m,i,x,y,k,j,top,s:longint;
     ch:string;

 procedure swap(var x,y:longint);
 var t:longint;
 begin
  t:=x; x:=y; y:=t;
 end;

 function isroot(x:longint):boolean;
 begin
  if (t[fa[x],]<>x)and(t[fa[x],]<>x) then exit(true);
  exit(false);
 end;

 procedure pushdown(x:longint);
 var l,r:longint;
 begin
  l:=t[x,]; r:=t[x,];
  if rev[x]> then
  begin
   rev[x]:=rev[x] xor ; rev[l]:=rev[l] xor ; rev[r]:=rev[r] xor ;
   swap(t[x,],t[x,]);
  end;
 end;

 procedure rotate(x:longint);
 var y,z,l,r:longint;
 begin
  y:=fa[x]; z:=fa[y];
  if t[y,]=x then l:=
   else l:=;
  r:=l xor ;
  while not isroot(y) do
  begin
   if t[z,]=y then t[z,]:=x
    else t[z,]:=x;
  end;
  fa[x]:=z; fa[y]:=x; fa[t[x,r]]:=y;
  t[y,l]:=t[x,r]; t[x,r]:=y;
 end;

 procedure splay(x:longint);
 var y,z,k:longint;
 begin
  inc(top); q[top]:=x;
  k:=x;
  while not isroot(k) do
  begin
   inc(top); q[top]:=fa[k];
   k:=fa[k];
  end;

  while top> do
  begin
   pushdown(q[top]);
   dec(top);
  end;

  while not isroot(x) do
  begin
   y:=fa[x]; z:=fa[y];
   if not isroot(y) then
   begin
    if (t[y,]=x)xor(t[z,]=y) then rotate(x)
     else rotate(y);
   end;
   rotate(x);
  end;
 end;

 procedure access(x:longint);
 var last:longint;
 begin
  last:=;
  while x> do
  begin
   splay(x); t[x,]:=last;
   last:=x; x:=fa[x];
  end;
 end;

 procedure makeroot(x:longint);
 begin
  access(x); splay(x); rev[x]:=rev[x] xor ;
 end;

 procedure link(x,y:longint);
 begin
  makeroot(x); fa[x]:=y;
 end;

 procedure cut(x,y:longint);
 begin
  makeroot(x); access(y); splay(y); t[y,]:=; fa[x]:=;
 end;

 function findroot(x:longint):longint;
 var k:longint;
 begin
  access(x); splay(x);
  k:=x;
  while t[k,]<> do k:=t[k,];
  exit(k);
 end;

 begin
  assign(input,'bzoj2049.in'); reset(input);
  assign(output,'bzoj2049.out'); rewrite(output);
  readln(n,m);
  for i:= to m do
  begin
   readln(ch); k:=length(ch); s:=; x:=; y:=; j:=;
   repeat
    if (ch[j]=' ') then
    begin
     inc(s);
     while ch[j]=' ' do inc(j);
     continue;
    end;
    if (ch[j]>='')and(ch[j]<='') then
    begin
     case s of
      :x:=x*+ord(ch[j])-ord('');
      :y:=y*+ord(ch[j])-ord('');
     end;
    end;
    inc(j);
   until j>k;
   case ch[] of
    'C':link(x,y);
    'D':cut(x,y);
    'Q':
     if findroot(x)<>findroot(y) then writeln('No')
      else writeln('Yes');
   end;
  end;

  close(input);
  close(output);
 end.