【BZOJ3295】动态逆序对（BIT套动态加点线段树）

题意：对于序列A，它的逆序对数定义为满足i<j，且A_i>A_j的数对(i,j)的个数。

给1到n的一个排列，按照某种顺序依次删除m个元素，你的任务是在每次删除一个元素之前统计整个序列的逆序对数。

N<=100000 M<=50000

思路：树套树即可，第i棵树维护前i位中1-j的个数，支持区间求和

外面要套BIT来实现区间加

本来维护剩下的数就可以，但初始化可能要开(n log^2 n) 个个节点，这题又只有128MB……

所以只能维护删除的数了，如果256MB乱搞就行了……

 var t:array[..,..]of longint;
     sum:array[..]of int64;
     root,a,b,a1,a2,c:array[..]of longint;
     n,m,i,x,cnt,l,r,j:longint;
     ans:int64;

 procedure pushup(x:longint);
 begin
  l:=t[x,]; r:=t[x,];
  sum[x]:=sum[l]+sum[r];
 end;

 procedure update(l,r,x,v:longint;var p:longint);
 var mid:longint;
 begin
  if p= then
  begin
   inc(cnt); p:=cnt;
  end;
  if l=r then
  begin
   sum[p]:=sum[p]+v; exit;
  end;
  mid:=(l+r)>>;
  if x<=mid then update(l,mid,x,v,t[p,]);
  if x>mid then update(mid+,r,x,v,t[p,]);
  pushup(p);
 end;

 function query(l,r,x,y,p:longint):int64;
 var mid:longint;
 begin
  if p= then exit();
  if (l>=x)and(r<=y) then exit(sum[p]);
  mid:=(l+r)>>;
  query:=;
  if x<=mid then query:=query+query(l,mid,x,y,t[p,]);
  if y>mid then query:=query+query(mid+,r,x,y,t[p,]);
  pushup(p);
 end;

 function lowbit(x:longint):longint;
 begin
  exit(x and (-x));
 end;

 procedure change(i,j,v:longint);
 var k:longint;
 begin
  k:=i;
  while k<=n do
  begin
   update(,n,j,v,root[k]);
   k:=k+lowbit(k);
  end;
 end;

 function getsum(x:longint):longint;
 begin
  getsum:=;
  while x> do
  begin
   getsum:=getsum+c[x];
   x:=x-lowbit(x);
  end;
 end;

 function ask(i,j,x,y:longint):int64;
 var k:longint;
 begin
  ask:=;
  if i>j then exit();
  if x>y then exit();
  k:=j;
  while k> do
  begin
   ask:=ask+query(,n,x,y,root[k]);
   k:=k-lowbit(k);
  end;
  k:=i-;
  while k> do
  begin
   ask:=ask-query(,n,x,y,root[k]);
   k:=k-lowbit(k);
  end;
 end;

 begin
  assign(input,'bzoj3295.in'); reset(input);
  assign(output,'bzoj3295.out'); rewrite(output);
  readln(n,m);

  for i:= to n do
  begin
   read(a[i]);
   b[a[i]]:=i;
   a1[i]:=getsum(n)-getsum(a[i]);
   ans:=ans+a1[i];
   j:=a[i];
   while j<=n do
   begin
    inc(c[j]);
    j:=j+lowbit(j);
   end;
  end;

  fillchar(c,sizeof(c),);
  for i:=n downto  do
  begin
   a2[i]:=getsum(a[i]-);
   j:=a[i];
   while j<=n do
   begin
    inc(c[j]);
    j:=j+lowbit(j);
   end;
  end;

 { for i:=1 to n do write(a1[i],' ');
  writeln;
  for i:=1 to n do write(a2[i],' ');  }
  for i:= to m do
  begin
   readln(x);
   writeln(ans);
   j:=b[x];
   ans:=ans-(a1[j]+a2[j]-ask(,j-,x+,n)-ask(j+,n,,x-));
   change(j,x,);
  end;

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