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[0..100010*4,0..1] of node;
    rev:array[0..100010*4] of boolean;
    a:array[0..100010] of longint;
    q:array[0..100010] 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,1],a[l]);
      put(tree[i,0],-a[l]);
      for j:=0 to 1 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 1;
      build(i*2,l,m);
      build(i*2+1,m+1,r);
      update(tree[i,1],tree[i*2,1],tree[i*2+1,1]);
      update(tree[i,0],tree[i*2,0],tree[i*2+1,0]);
    end;
  end;

procedure change(i:longint);
  begin
    swap(tree[i,0],tree[i,1]);
    rev[i]:=not rev[i];
  end;

procedure push(i:longint);
  begin
    rev[i]:=false;
    change(i*2);
    change(i*2+1);
  end;

procedure work(i,l,r:longint);
  var m:longint;
  begin
    if l=r then
    begin
      put(tree[i,1],y);
      put(tree[i,0],-y);
    end
    else begin
      m:=(l+r) shr 1;
      if rev[i] then push(i);
      if x<=m then work(i*2,l,m)
      else work(i*2+1,m+1,r);
      update(tree[i,1],tree[i*2,1],tree[i*2+1,1]);
      update(tree[i,0],tree[i*2,0],tree[i*2+1,0]);
    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,1])
    else begin
      m:=(l+r) shr 1;
      if rev[i] then push(i);
      if y<=m then exit(ask(i*2,l,m));
      if x>m then exit(ask(i*2+1,m+1,r));
      s1:=ask(i*2,l,m);
      s2:=ask(i*2+1,m+1,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 1;
      if rev[i] then push(i);
      if x<=m then rever(i*2,l,m,x,y);
      if y>m then rever(i*2+1,m+1,r,x,y);
      update(tree[i,1],tree[i*2,1],tree[i*2+1,1]);
      update(tree[i,0],tree[i*2,0],tree[i*2+1,0]);
    end;
  end;

begin
  readln(n);
  for i:=1 to n do
    read(a[i]);
  build(1,1,n);
  readln(m);
  for i:=1 to m do
  begin
    read(ch);
    if ch=1 then
    begin
      readln(x,y,k);
      ans:=0;
      t:=0;
      for j:=1 to k do
      begin
        c:=ask(1,1,n);
        if c.mx>0 then ans:=ans+c.mx
        else break;
        rever(1,1,n,c.pb,c.pe);
        inc(t);
        q[t]:=c;
      end;
      for j:=t downto 1 do
        rever(1,1,n,q[j].pb,q[j].pe);
      writeln(ans);
    end
    else begin
      readln(x,y);
      work(1,1,n);
    end;
  end;
end.