bzoj 2836 树链剖分

题意:对一棵n个节点且每个节点初始值都为0的树,进行两个操作:

1、Add x y z 表示x到y的路径上的点的点权都+z

2、Query x 表示询问以x为根的子树的点权和

树链剖分裸题,mdzz,8A....

对于修改x到y的路径上的点的点权,注意x到y并不是一段连续的区间,需要跳到同一条链上才能用线段树进行区间修改

对于查询,由于以x为根的子树在线段树中是一段连续的区间[num[x],num[x]+size[x]-1],并不需要把左右端点跳到同一条链上,不然会多算...(mdzz,当时没多想,直接复制的wawawa....= =)

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

var
    n,m,l,x,y,z,tot     :longint;
    i                   :longint;
    ch                  :char;
    s                   :string;
    last,num,father     :array[0..100010] of longint;
    size,max_son,d,top  :array[0..100010] of longint;
    vis                 :array[0..100010] of boolean;
    pre,other           :array[0..200010] of longint;
    t                   :array[0..400010] of rec;
procedure connect(x,y:longint);
begin
   inc(l);
   pre[l]:=last[x];
   last[x]:=l;
   other[l]:=y;
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 exit;
   mid:=(l+r)>>1;
   build(2*x,l,mid); build(2*x+1,mid+1,r);
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(x*2,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+1].sum+t[2*x].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,z:longint);
var
    c:longint;
begin
   if d[x]>d[y] then
   begin
      c:=x; x:=y; y:=c;
   end;
   while (d[x]top[y]) do
   begin
      change(1,num[top[y]],num[y],z);
      change(1,num[top[x]],num[x],z);
      x:=father[top[x]];
      y:=father[top[y]];
   end;
   if num[x]' ') do
      begin
         s:=s+ch;
         read(ch);
      end;
      if (s='Add') then
      begin
         readln(x,y,z);
         inc(x); inc(y);
         change1(x,y,z);
      end else
      begin
         readln(x);
         inc(x);
         writeln(get_sum(1,num[x],num[x]+size[x]-1));
      end;
   end;
end.

——by Eirlys

bzoj 2836 树链剖分_第1张图片

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