bzoj1109

我们设f[i]为保留第i个木块最多的符合未知数

显然f[i]=max(f[j])+1 满足i>j a[i]>a[j] i-j>=a[i]-a[j]

我们把最后一个式子变成a[i]-i<=a[j]-j

三元关系很不好弄对吧,但仔细观察可知，由最后两个式子一定可以推出i>j

那不就水了，排序树状数组即可

 var f,a,b:array[..] of longint;
     c:array[..] of longint;
     i,n,ans:longint;

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

 function lowbit(x:longint):longint;
   begin
     exit(x and (-x));
   end;

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

 procedure add(x,w:longint);
   begin
     while x<=n do
     begin
       c[x]:=max(c[x],w);
       x:=x+lowbit(x);
     end;
   end;

 function ask(x:longint):longint;
   begin
     ask:=;
     while x> do
     begin
       ask:=max(ask,c[x]);
       x:=x-lowbit(x);
     end;
   end;

 procedure sort(l,r:longint);
   var y,i,j,x:longint;
   begin
     i:=l;
     j:=r;
     x:=a[(l+r) shr ];
     y:=b[(l+r) shr ];
     repeat
       while (b[i]>y) or (b[i]=y) and (a[i]<x) do inc(i);
       while (y>b[j]) or (b[j]=y) and (x<a[j]) do dec(j);
       if not(i>j) then
       begin
         swap(a[i],a[j]);
         swap(b[i],b[j]);
         inc(i);
         dec(j);
       end;
     until i>j;
     if l<j then sort(l,j);
     if i<r then sort(i,r);
   end;

 begin
   readln(n);
   for i:= to n do
   begin
     read(a[i]);
     b[i]:=a[i]-i;
   end;
   sort(,n);
   for i:= to n do
   begin
     if b[i]> then continue;
     f[i]:=ask(a[i]-)+;
     ans:=max(f[i],ans);
     add(a[i],f[i]);
   end;
   writeln(ans);
 end.