【BZOJ4476&JSOI2015】送礼物（二分，RMQ）

ANS明显是有二分性的

二分答案,设二分值为b
M(i,j)−m(i,j)j−i+k>b

显然当l<长度<r时，一端是最小值，一端是最大值。

等于l或r的时候因为可能不满足以上性质，所以RMQ暴力O(nlogn)做。

a[i]−a[j]>b∗j−b∗i+b∗k 或 a[j]−a[i]>b∗j−b∗i+b∗k
那么
(a[i]+b∗i)−(a[j]+b∗j)>b∗k 或 (−a[i]+b∗i)−(−a[j]+b∗j)>b∗k

是一个单调队列的样子

 var f1,f2:array[..,..]of extended;
     q:array[..]of longint;
     a,e:array[..]of extended;
     mi:array[..]of int64;
     log:array[..]of longint;
     cas,i,v,j:longint;
     n,l1,r1,lg:int64;
     l,r,mid,k:extended;

 function max(x,y:extended):extended;
 begin
  if x>y then exit(x);
  exit(y);
 end;

 function min(x,y:extended):extended;
 begin
  if x<y then exit(x);
  exit(y);
 end;

 function querymin(x,y:longint):extended;
 var len:longint;
 begin
  len:=log[y-x+];
  exit(min(f1[x,len],f1[y-mi[len]+,len]));
 end;

 function querymax(x,y:longint):extended;
 var len:longint;
 begin
  len:=log[y-x+];
  exit(max(f2[x,len],f2[y-mi[len]+,len]));
 end;

 procedure build1;
 var i,j:longint;
 begin
  for i:= to lg do
   for j:= to n-mi[i]+ do f1[j,i]:=min(f1[j,i-],f1[j+mi[i-],i-]);
 end;

 procedure build2;
 var i,j:longint;
 begin
  for i:= to lg do
   for j:= to n-mi[i]+ do f2[j,i]:=max(f2[j,i-],f2[j+mi[i-],i-]);
 end;

 function isok(b:extended):boolean;
 var h,w,i:longint;
 begin
  for i:= to n do e[i]:=a[i]+b*i;
  h:=; w:=;
  q[]:=n;
  for i:=n-l1+ downto  do
  begin
   while (h<=w)and(q[h]>i+r1-) do inc(h);
   if e[i]-e[q[h]]>=b*k then exit(true);
   while (h<=w)and(e[q[w]]>=e[i+l1-]) do dec(w);
   inc(w); q[w]:=i+l1-;
  end;
  for i:= to n do e[i]:=-a[i]+b*i;
  h:=; w:=;
  q[]:=n;
  for i:=n-l1+ downto  do
  begin
   while (h<=w)and(q[h]>i+r1-) do inc(h);
   if e[i]-e[q[h]]>=b*k then exit(true);
   while (h<=w)and(e[q[w]]>=e[i+l1-]) do dec(w);
   inc(w); q[w]:=i+l1-;
  end;
  exit(false);
 end;

 begin
  assign(input,'gift.in'); reset(input);
  assign(output,'gift.out'); rewrite(output);
  readln(cas);
  mi[]:=;
  for i:= to  do mi[i]:=mi[i-]*;
  for i:= to  do
   for j:=mi[i] to mi[i+]- do log[j]:=i;
  for v:= to cas do
  begin
   readln(n,k,l1,r1);
   for i:= to n do read(a[i]);
   l:=; r:=;
   for i:= to n do
   begin
    f1[i,]:=a[i];
    f2[i,]:=a[i];
   end;
   lg:=log[r1];
   build1; build2;
   for i:= to n-l1+ do l:=max(l,(querymax(i,i+l1-)-querymin(i,i+l1-))/(l1-+k));
   for i:= to n-r1+ do l:=max(l,(querymax(i,i+r1-)-querymin(i,i+r1-))/(r1-+k));
   l1:=l1+; r1:=r1-;
   if l1<=r1 then
    while r-l>1e-6 do
    begin
     mid:=(r+l)/;
     if isok(mid) then l:=mid
      else r:=mid;
    end;
   writeln(l::);
  end;
  close(input);
  close(output);
 end.