【BZOJ3223】文艺平衡树（splay）

题意：维护数列的翻转

n<=100000

思路：裸splay，仅维护rever

var t:array[0..200000,0..1]of longint;
    size,rev,b,fa,a:array[0..200000]of longint;
    n,m,i,x,y,root:longint;

procedure pushup(p:longint);
begin
 size[p]:=size[t[p,0]]+size[t[p,1]]+1;
end;

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

procedure pushdown(p:longint);
var l,r:longint;
begin
 l:=t[p,0]; r:=t[p,1];
 if rev[p]=1 then
 begin
  rev[p]:=rev[p] xor 1; rev[l]:=rev[l] xor 1; rev[r]:=rev[r] xor 1;
  swap(t[p,0],t[p,1]);
 end;
end;

procedure rotate(x:longint;var k:longint);
var y,z,l,r:longint;
begin
 y:=fa[x]; z:=fa[y];
 if t[y,0]=x then l:=0
  else l:=1;
 r:=l xor 1;
 if y<>k then
 begin
  if t[z,0]=y then t[z,0]:=x
   else t[z,1]:=x;
 end
  else k:=x;
 fa[x]:=z; fa[y]:=x; fa[t[x,r]]:=y;
 t[y,l]:=t[x,r]; t[x,r]:=y;
 pushup(y);
 pushup(x);
end;

procedure splay(x:longint;var k:longint);
var y,z:longint;
begin
 while x<>k do
 begin
  y:=fa[x]; z:=fa[y];
  if y<>k then
  begin
   if (t[y,0]=x)xor(t[z,0]=y) then rotate(x,k)
    else rotate(y,k)
  end
   else k:=x;
  rotate(x,k);
 end;
end;

procedure build(l,r,x:longint);
var mid,now:longint;
begin
 if l>r then exit;
 mid:=(l+r)>>1; now:=mid;
 if l=r then
 begin
  size[now]:=1;
 end
  else
  begin
   build(l,mid-1,mid);
   build(mid+1,r,mid);
  end;
 b[now]:=a[mid]; fa[now]:=x;
 pushup(now);
 if mid>=x then t[x,1]:=now
  else t[x,0]:=now;
end;

function findkth(x:longint):longint;
var k,tmp:longint;
begin
 k:=root;
 while k<>0 do
 begin
  pushdown(k);
  tmp:=size[t[k,0]]+1;
  if tmp=x then exit(k)
   else if tmp>x then k:=t[k,0]
    else
    begin
     x:=x-tmp;
     k:=t[k,1];
    end;
 end;
end;

procedure rever(x,y:longint);
var i,j:longint;
begin
 i:=findkth(x); j:=findkth(y+2);
 splay(i,root);
 splay(j,t[i,1]);
 x:=t[j,0];
 rev[x]:=rev[x] xor 1;
end;

begin
 assign(input,'bzoj3223.in'); reset(input);
 assign(output,'bzoj3223.out'); rewrite(output);
 readln(n,m);
 a[1]:=-maxlongint; a[n+2]:=maxlongint;
 for i:=2 to n+1 do a[i]:=i;
 build(1,n+2,0); root:=(n+3)>>1;
 for i:=1 to m do
 begin
  readln(x,y);
  rever(x,y);
 end;
 for i:=2 to n+1 do
  write(b[findkth(i)]-1,' ');
 close(input);
 close(output);
end.