bzoj3747

经典题，记录每个位置对应数下次出现的位置next[i]

每个位置维护当前左端点下到这个位置的和

随着左端点的右移一位到i+1，对[i+1,next[i]-1] 的影响是-a[i], [next[i],next[next[i]]-1]的影响是a[i]

线段树维护，pascal过不去系列

UPD:现在时限开大能过去了

 var lazy,tree:array[..*] of int64;
     s:array[..] of int64;
     next,last,w,a:array[..] of longint;
     i,n,m:longint;
     ans:int64;

 function max(a,b:int64):int64;
   begin
     if a>b then exit(a) else exit(b);
   end;

 procedure ins(i:longint;z:int64);
   begin
     tree[i]:=tree[i]+z;
     lazy[i]:=lazy[i]+z;
   end;

 procedure push(i:longint);
   begin
     ins(i*,lazy[i]);
     ins(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 ins(i,z)
     else begin
       m:=(l+r) shr ;
       if lazy[i]<> then push(i);
       if x<=m then add(i*,l,m,x,y,z);
       if y>m then add(i*+,m+,r,x,y,z);
       tree[i]:=max(tree[i*],tree[i*+]);
     end;
   end;

 begin

   readln(n,m);
   for i:= to n do
     read(a[i]);
   for i:= to m do
   begin
     read(w[i]);
     last[i]:=n+;
   end;
   for i:=n downto  do
   begin
     if last[a[i]]=n+ then s[i]:=s[i+]+w[a[i]]
     else s[i]:=s[i+];
     next[i]:=last[a[i]];
     last[a[i]]:=i;
   end;
   for i:= to m do
     if last[i]> then add(,,n,last[i],next[last[i]]-,w[i]);

   for i:= to n do
   begin
     if s[i]<=ans then break;
     ans:=max(ans,tree[]);
     add(,,n,i,next[i]-,-w[a[i]]);
     if next[i]<=n then add(,,n,next[i],next[next[i]]-,w[a[i]]);
   end;
   writeln(ans);
 end.