bzoj3262

三维裸的做法是一维排序，剩下树套树，可我好像还没写过树套树
先说cdq分治吧，先对一维排序，相当于原来修改询问里的时间线
在这上面分治、划分，计算前半部分对后半部分的影响，显然可以按第二维的顺序维护树状数组

 type node=record
        a,b,c,s,p:longint;
      end;

 var a,b,q:array[..] of node;
     ans,c:array[..] of longint;
     n,m,i,t,j:longint;

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

 function cmp(a,b:node):boolean;
   begin
     if a.a<b.a then exit(true);
     if a.a>b.a then exit(false);
     if (a.b<b.b) or (a.b=b.b) and (a.c<b.c) then exit(true);
     exit(false);
   end;

 procedure sorta(l,r:longint);
   var i,j:longint;
       x:node;
   begin
     i:=l;
     j:=r;
     x:=a[(l+r) shr ];
     repeat
       while cmp(a[i],x) do inc(i);
       while cmp(x,a[j]) do dec(j);
       if not(i>j) then
       begin
         swap(a[i],a[j]);
         inc(i);
         dec(j);
       end;
     until i>j;
     if l<j then sorta(l,j);
     if i<r then sorta(i,r);
   end;

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

 procedure add(x,w:longint);
   begin
     while x<=m do
     begin
       inc(c[x],w);
       x:=x+lowbit(x);
     end;
   end;

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

 procedure cdq(l,r:longint);
   var m,i,j,l1,l2:longint;
   begin
     if l=r then exit;
     m:=(l+r) shr ;
     cdq(l,m);
     cdq(m+,r);
     j:=l;
     for i:=m+ to r do
     begin
       while (j<=m) and (b[j].b<=b[i].b) do
       begin
         add(b[j].c,b[j].p);
         inc(j);
       end;
       inc(b[i].s,ask(b[i].c));
     end;
     for i:=l to j- do
       add(b[i].c,-b[i].p);
     l1:=l; l2:=m+;
     for i:=l to r do
       if ((l1<=m) and (b[l1].b<b[l2].b)) or (l2>r) then
       begin
         q[i]:=b[l1];
         inc(l1);
       end
       else begin
         q[i]:=b[l2];
         inc(l2);
       end;
     for i:=l to r do
       b[i]:=q[i];
   end;

 begin
   readln(n,m);
   for i:= to n do
     readln(a[i].a,a[i].b,a[i].c);
   sorta(,n);
   i:=;
   while i<n do
   begin
     inc(i);
     j:=i+;
     while (a[j].a=a[i].a) and (a[j].b=a[i].b) and (a[j].c=a[i].c) do inc(j);  //这步不能少，因为要保证后半部分序列对前半部分没有影响
     inc(t);
     b[t]:=a[i];
     b[t].p:=j-i;
     i:=j-;
   end;
   cdq(,t);
   for i:= to t do
     inc(ans[b[i].s+b[i].p-],b[i].p);
   for i:= to n- do
     writeln(ans[i]);
 end.