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

1 type node=record  2        a,b,c,s,p:longint;  3      end;  4   5 var a,b,q:array[0..100010] of node;  6     ans,c:array[0..200010] of longint;  7     n,m,i,t,j:longint;  8   9 procedure swap(var a,b:node); 10   var c:node; 11   begin 12     c:=a; 13     a:=b; 14     b:=c; 15   end; 16  17 function cmp(a,b:node):boolean; 18   begin 19     if a.a
      
       b.a then exit(false); 21     if (a.b
       
        j) then 36       begin 37         swap(a[i],a[j]); 38         inc(i); 39         dec(j); 40       end; 41     until i>j; 42     if l
        
         0 do 64     begin 65       ask:=ask+c[x]; 66       x:=x-lowbit(x); 67     end; 68   end; 69  70 procedure cdq(l,r:longint); 71   var m,i,j,l1,l2:longint; 72   begin 73     if l=r then exit; 74     m:=(l+r) shr 1; 75     cdq(l,m); 76     cdq(m+1,r); 77     j:=l; 78     for i:=m+1 to r do 79     begin 80       while (j<=m) and (b[j].b<=b[i].b) do 81       begin 82         add(b[j].c,b[j].p); 83         inc(j); 84       end; 85       inc(b[i].s,ask(b[i].c)); 86     end; 87     for i:=l to j-1 do 88       add(b[i].c,-b[i].p); 89     l1:=l; l2:=m+1; 90     for i:=l to r do 91       if ((l1<=m) and (b[l1].b
         
          r) then 92       begin 93         q[i]:=b[l1]; 94         inc(l1); 95       end 96       else begin 97         q[i]:=b[l2]; 98         inc(l2); 99       end;100     for i:=l to r do101       b[i]:=q[i];102   end;103 104 begin105   readln(n,m);106   for i:=1 to n do107     readln(a[i].a,a[i].b,a[i].c);108   sorta(1,n);109   i:=0;110   while i