bzoj2482

还是像以前那样维护下次出现位置，计算影响

其实不难，思维盲点，受到做最大子段和的影响

其实这里可以直接维护当前每个位置的子段和，再记录一个历史最大和

当然tag也需要记录当前tag和历史（距离上次push）最大累加

 type node=record
        x,y,id:longint;
      end;

 var lazy,tree:array[..*,..] of longint;
     ans,next,a:array[..] of longint;
     last:array[-..] of longint;
     q:array[..] of node;
     i,j,n,m:longint;

 function max(a,b:longint):longint;
   begin
     if a>b then exit(a) else exit(b);
   end;

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

 procedure sort(l,r:longint);
   var i,j,x:longint;
   begin
     i:=l;
     j:=r;
     x:=q[(l+r) shr ].x;
     repeat
       while q[i].x>x do inc(i);
       while x>q[j].x do dec(j);
       if 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 get(i,x0,x1:longint);
   begin
     tree[i,]:=max(tree[i,],tree[i,]+max(x0,x1));
     lazy[i,]:=max(lazy[i,],lazy[i,]+max(x0,x1));
     inc(lazy[i,],x1);
     inc(tree[i,],x1);
   end;

 procedure update(i:longint);
   begin
     tree[i,]:=max(tree[i*,],tree[i*+,]);
     tree[i,]:=max(tree[i*,],tree[i*+,]);
   end;

 procedure push(i:longint);
   begin
     if (lazy[i,]=) and (lazy[i,]=) then exit;
     get(i*,lazy[i,],lazy[i,]);
     get(i*+,lazy[i,],lazy[i,]);
     lazy[i,]:=;
     lazy[i,]:=;
   end;

 procedure add(i,l,r,x,y,z:longint);
   var m:longint;
   begin
     if (x<=l) and (y>=r) then get(i,,z)
     else begin
       m:=(l+r) shr ;
       push(i);
       if x<=m then add(i*,l,m,x,y,z);
       if y>m then add(i*+,m+,r,x,y,z);
       update(i);
     end;
   end;

 function ask(i,l,r,x:longint):longint;
   var m:longint;
   begin
     if l=r then exit(tree[i,])
     else begin
       m:=(l+r) shr ;
       push(i);
       if x<=m then exit(ask(i*,l,m,x))
       else exit(max(tree[i*,],ask(i*+,m+,r,x)));
     end;
   end;

 begin
   readln(n);
   for i:= to n do
     read(a[i]);
   for i:=n downto  do
   begin
     next[i]:=last[a[i]];
     last[a[i]]:=i;
   end;
   readln(m);
   for i:= to m do
   begin
     readln(q[i].x,q[i].y);
     q[i].id:=i;
   end;
   sort(,m);
   j:=;
   for i:=n downto  do
   begin
     if next[i]= then next[i]:=n+;
     add(,,n,i,next[i]-,a[i]);
     while (j<=m) and (q[j].x=i) do
     begin
       ans[q[j].id]:=ask(,,n,q[j].y);
       inc(j);
     end;
     if j=m+ then break;
   end;
   for i:= to m do
     writeln(ans[i]);
 end.