bzoj3252

简单题，每次取出最长链，然后对于练上每个点x，终点在其子树内的链都要减去a[x]

这显然可以用dfs序+线段树维护

显然每个点只要删一次即可，复杂度是O(nlogn)

 type node=record
        s,lx,rx,mx,lp,rp,pb,pe:longint;
      end;

 var tree:array[..*,..] of node;
     rev:array[..*] of boolean;
     a:array[..] of longint;
     q:array[..] of node;
     j,t,ans,i,n,m,x,y,k,ch:longint;
     c:node;

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

 procedure put(var a:node; x:longint);
   begin
     a.lx:=x;
     a.rx:=x;
     a.mx:=x;
     a.s:=x;
   end;

 procedure update(var c:node; a,b:node);
   begin
     c.s:=a.s+b.s;
     c.lx:=a.lx; c.lp:=a.lp;
     if a.s+b.lx>c.lx then
     begin
       c.lx:=a.s+b.lx;
       c.lp:=b.lp;
     end;
     c.rx:=b.rx; c.rp:=b.rp;
     if b.s+a.rx>c.rx then
     begin
       c.rx:=b.s+a.rx;
       c.rp:=a.rp;
     end;

     c.mx:=a.rx+b.lx; c.pb:=a.rp; c.pe:=b.lp;
     if c.mx<a.mx then
     begin
       c.mx:=a.mx;
       c.pb:=a.pb;
       c.pe:=a.pe;
     end;
     if c.mx<b.mx then
     begin
       c.mx:=b.mx;
       c.pb:=b.pb;
       c.pe:=b.pe;
     end;
   end;

 procedure build(i,l,r:longint);
   var m,j:longint;
   begin
     if l=r then
     begin
       put(tree[i,],a[l]);
       put(tree[i,],-a[l]);
       for j:= to  do
       begin
         tree[i,j].lp:=l;
         tree[i,j].rp:=l;
         tree[i,j].pb:=l;
         tree[i,j].pe:=l;
       end;
     end
     else begin
       m:=(l+r) shr ;
       build(i*,l,m);
       build(i*+,m+,r);
       update(tree[i,],tree[i*,],tree[i*+,]);
       update(tree[i,],tree[i*,],tree[i*+,]);
     end;
   end;

 procedure change(i:longint);
   begin
     swap(tree[i,],tree[i,]);
     rev[i]:=not rev[i];
   end;

 procedure push(i:longint);
   begin
     rev[i]:=false;
     change(i*);
     change(i*+);
   end;

 procedure work(i,l,r:longint);
   var m:longint;
   begin
     if l=r then
     begin
       put(tree[i,],y);
       put(tree[i,],-y);
     end
     else begin
       m:=(l+r) shr ;
       if rev[i] then push(i);
       if x<=m then work(i*,l,m)
       else work(i*+,m+,r);
       update(tree[i,],tree[i*,],tree[i*+,]);
       update(tree[i,],tree[i*,],tree[i*+,]);
     end;
   end;

 function ask(i,l,r:longint):node;
   var m:longint;
       s,s1,s2:node;
   begin
     if (x<=l) and (y>=r) then exit(tree[i,])
     else begin
       m:=(l+r) shr ;
       if rev[i] then push(i);
       if y<=m then exit(ask(i*,l,m));
       if x>m then exit(ask(i*+,m+,r));
       s1:=ask(i*,l,m);
       s2:=ask(i*+,m+,r);
       update(s,s1,s2);
       exit(s);
     end;
   end;

 procedure rever(i,l,r,x,y:longint);
   var m:longint;
   begin
     if (x<=l) and (y>=r) then change(i)
     else begin
       m:=(l+r) shr ;
       if rev[i] then push(i);
       if x<=m then rever(i*,l,m,x,y);
       if y>m then rever(i*+,m+,r,x,y);
       update(tree[i,],tree[i*,],tree[i*+,]);
       update(tree[i,],tree[i*,],tree[i*+,]);
     end;
   end;

 begin
   readln(n);
   for i:= to n do
     read(a[i]);
   build(,,n);
   readln(m);
   for i:= to m do
   begin
     read(ch);
     if ch= then
     begin
       readln(x,y,k);
       ans:=;
       t:=;
       for j:= to k do
       begin
         c:=ask(,,n);
         if c.mx> then ans:=ans+c.mx
         else break;
         rever(,,n,c.pb,c.pe);
         inc(t);
         q[t]:=c;
       end;
       for j:=t downto  do
         rever(,,n,q[j].pb,q[j].pe);
       writeln(ans);
     end
     else begin
       readln(x,y);
       work(,,n);
     end;
   end;
 end.