bzoj3626

百度空间马上要下架的说，赶快把最后一点题解补完，然后搬家
这是一道不错的题，首先注意询问是满足区间减法的，我们把他变成前缀和表示
设我们询问[1,r]中的点和z的LCA深度和，假设我们确定一个根，不难发现一个有趣的事情
点z和点i的LCA深度=z和i到根公共路径（LCA到根的路径）上点的个数！
也就是说，当我们查询时，我们只要知道，[1,r]上的点到根路径上的点在z到根上出现的次数和
所以不难想到离线的做法，按照编号的顺序，每次做到点i，就把点i到根路径上点权值+1
询问的时候我们只要查询z到根路径上点权值和即可，显然我们可以用树链剖分+线段树维护

 const mo=;
 type node=record
        po,next:longint;
      end;
      que=record
        x,y,p,op:longint;
      end;

 var p,c,top,ans,fa,size:array[..] of longint;
     tree,lazy:array[..*] of longint;
     q:array[..] of que;
     e:array[..] of node;
     t,l,r,x,i,j,n,m:longint;

 procedure swap(var a,b:que);
   var c:que;
   begin
     c:=a;
     a:=b;
     b:=c;
   end;

 procedure add(x,y:longint);
   begin
     e[i].po:=y;
     e[i].next:=p[x];
     p[x]:=i;
   end;

 procedure dfs1(x:longint);
   var i,y:longint;
   begin
     size[x]:=;
     i:=p[x];
     while i<> do
     begin
       y:=e[i].po;
       fa[y]:=x;
       dfs1(y);
       size[x]:=size[x]+size[y];
       i:=e[i].next;
     end;
   end;

 procedure dfs2(x:longint);
   var i,y,q:longint;
   begin
     inc(t);
     c[x]:=t;
     q:=n+;
     i:=p[x];
     while i<> do
     begin
       y:=e[i].po;
       if size[y]>size[q] then q:=y;
       i:=e[i].next;
     end;
     if q<>n+ then
     begin
       top[q]:=top[x];
       dfs2(q);
     end;
     i:=p[x];
     while i<> do
     begin
       y:=e[i].po;
       if y<>q then
       begin
         top[y]:=y;
         dfs2(y);
       end;
       i:=e[i].next;
     end;
   end;

 procedure sort(l,r:longint);
   var i,j,y:longint;
   begin
     i:=l;
     j:=r;
     y:=q[(l+r) shr ].y;
     repeat
       while q[i].y<y do inc(i);
       while y<q[j].y do dec(j);
       if not(i>j) then
       begin
         swap(q[i],q[j]);
         inc(i);
         dec(j);
       end;
     until i>j;
     if l<j then sort(l,j);
     if i<r then sort(i,r);
   end;

 procedure push(i,l,r:longint);
   var m:longint;
   begin
     m:=(l+r) shr ;
     tree[i*]:=tree[i*]+lazy[i]*(m+-l);
     tree[i*+]:=tree[i*+]+lazy[i]*(r-m);
     inc(lazy[i*],lazy[i]);
     inc(lazy[i*+],lazy[i]);
     lazy[i]:=;
   end;

 procedure work(i,l,r,x,y:longint);
   var m:longint;
   begin
     if (x<=l) and (y>=r) then
     begin
       tree[i]:=tree[i]+r-l+;
       lazy[i]:=lazy[i]+;
     end
     else begin
       if lazy[i]> then push(i,l,r);
       m:=(l+r) shr ;
       if x<=m then work(i*,l,m,x,y);
       if y>m then work(i*+,m+,r,x,y);
       tree[i]:=(tree[i*]+tree[i*+]) mod mo;
     end;
   end;

 function getans(i,l,r,x,y:longint):longint;
   var m,s:longint;
   begin
     if (x<=l) and (y>=r) then exit(tree[i])
     else begin
       if lazy[i]> then push(i,l,r);
       m:=(l+r) shr ;
       s:=;
       if x<=m then s:=s+getans(i*,l,m,x,y);
       if y>m then s:=s+getans(i*+,m+,r,x,y);
       exit(s);
     end;
   end;

 procedure ins(x:longint);
   begin
     while top[x]<>- do
     begin
       work(,,n,c[top[x]],c[x]);
       x:=fa[top[x]];
     end;
     work(,,n,,c[x]);
   end;

 function ask(x:longint):longint;
   begin
     ask:=;
     while top[x]<>- do
     begin
       ask:=ask+getans(,,n,c[top[x]],c[x]);
       x:=fa[top[x]];
     end;
     ask:=(ask+getans(,,n,,c[x])) mod mo;
   end;

 begin
   readln(n,m);
   for i:= to n- do
   begin
     read(x);
     add(x,i);
   end;
   dfs1();
   top[]:=-;
   dfs2();
   t:=;
   for i:= to m do
   begin
     readln(l,r,x);
     inc(t);
     q[t].x:=x; q[t].y:=l-;
     q[t].p:=i; q[t].op:=-;
     inc(t);
     q[t].x:=x; q[t].y:=r;
     q[t].p:=i; q[t].op:=;
   end;
   sort(,t);
   j:=;
   for i:= to t do
   begin
     while (j<n) and (j<=q[i].y) do
     begin
       ins(j);
       inc(j);
     end;
     ans[q[i].p]:=(ans[q[i].p]+q[i].op*ask(q[i].x)+mo) mod mo;
   end;
   for i:= to m do
     writeln(ans[i]);
 end.