bzoj 3626 树链剖分+离线处理

题意：给出一个n个节点的有根树（编号为0到n-1，根节点为0）。

一个点的深度定义为这个节点到根的距离+1。

q个询问，每个询问给出区间[l,r]和z，求在[l,r]区间内的每个节点i与z的最近公共祖先的深度之和

好题...orz

并不会写题解...

const
        mo=201314;
type
        rec=record
            l,r:longint;
            sum,lazy:int64;
end;

type
        rec2=record
            p,num,z,op:longint;
end;

var
        n,m,l,ll,rr,z   :longint;
        i               :longint;
        now,tot,x,y     :longint;
        q               :array[0..100010] of rec2;
        last,size,top,d :array[0..50010] of longint;
        father,max_son  :array[0..50010] of longint;
        num             :array[0..50010] of longint;
        vis             :array[0..50010] of boolean;
        t               :array[0..200010] of rec;
        pre,other       :array[0..100010] of longint;
        ans             :array[0..50010] of int64;
procedure connect(x,y:longint);
begin
   inc(l);
   pre[l]:=last[x];
   last[x]:=l;
   other[l]:=y;
end;

procedure sort(l,r:longint);
var
        i,j,x:longint;
        y:rec2;
begin
   i:=l; j:=r; x:=q[(l+r)>>1].p;
   while (i<=j) do
   begin
      while q[i].px do dec(j);
      if (i<=j) then
      begin
         y:=q[i]; q[i]:=q[j]; q[j]:=y;
         inc(i); dec(j);
      end;
   end;
   if il then sort(l,j);
end;

procedure dfs(x:longint);
var
        p,q:longint;
begin
   size[x]:=1;
   q:=last[x];
   while (q<>0) do
   begin
      p:=other[q];
      if not vis[p] then
      begin
         vis[p]:=true;
         father[p]:=x;
         dfs(p);
         inc(size[x],size[p]);
         if size[max_son[x]]0 then
   begin
      vis[max_son[x]]:=true;
      make(max_son[x],t,depth);
   end;
   q:=last[x];
   while (q<>0) do
   begin
      p:=other[q];
      if not vis[p] and (p<>max_son[x]) then
      begin
         vis[p]:=true;
         make(p,p,depth+1);
      end;
      q:=pre[q];
   end;
end;

procedure build(x,l,r:longint);
var
        mid:longint;
begin
   t[x].l:=l; t[x].r:=r;
   if l=r then
   begin
      t[x].sum:=1;exit;
   end;
   mid:=(l+r)>>1;
   build(2*x,l,mid); build(2*x+1,mid+1,r);
   t[x].sum:=t[x*2].sum+t[2*x+1].sum;
end;

procedure update(x:longint);
begin
   inc(t[x].sum,t[x].lazy*int64(t[x].r-t[x].l+1));
   if t[x].l=t[x].r then
   begin
      t[x].lazy:=0;exit;
   end;
   inc(t[2*x].lazy,t[x].lazy);
   inc(t[2*x+1].lazy,t[x].lazy);
   t[x].lazy:=0;
end;

procedure change(x,l,r,z:longint);
var
        mid:longint;
begin
   if (t[x].l=l) and (t[x].r=r) then
   begin
      inc(t[x].lazy,z);exit;
   end;
   if t[x].lazy<>0 then update(x);
   mid:=(t[x].l+t[x].r)>>1;
   if r<=mid then change(2*x,l,r,z) else
     if l>mid then change(2*x+1,l,r,z) else
     begin
        change(2*x,l,mid,z);
        change(2*x+1,mid+1,r,z);
     end;
   t[x].sum:=t[2*x].sum+t[2*x+1].sum
            +t[2*x].lazy*int64(t[2*x].r-t[2*x].l+1)
            +t[2*x+1].lazy*int64(t[2*x+1].r-t[2*x+1].l+1);
end;

function get_sum(x,l,r:longint):int64;
var
        mid:longint;
begin
   if (t[x].lazy<>0) then update(x);
   if (t[x].l=l) and (t[x].r=r) then exit(t[x].sum);
   mid:=(t[x].l+t[x].r)>>1;
   if r<=mid then exit(get_sum(2*x,l,r)) else
     if l>mid then exit(get_sum(2*x+1,l,r)) else
       exit(get_sum(2*x,l,mid)+get_sum(2*x+1,mid+1,r));
end;

procedure change1(x,y:longint);
begin
   while (d[x]top[y]) do
   begin
      change(1,num[top[y]],num[y],1);
      change(1,num[top[x]],num[x],1);
      y:=father[top[y]];
      x:=father[top[x]];
   end;
   if num[x]top[y]) do
   begin
      inc(ans,get_sum(1,num[top[y]],num[y]));
      inc(ans,get_sum(1,num[top[x]],num[x]));
      y:=father[top[y]];
      x:=father[top[x]];
   end;
   if num[x]

——by Eirlys