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

1 var f,a,b:array[0..100010] of longint; 2     c:array[0..1000010] of longint; 3     i,n,ans:longint; 4  5 function max(a,b:longint):longint; 6   begin 7     if a>b then exit(a) else exit(b); 8   end; 9 10 function lowbit(x:longint):longint;11   begin12     exit(x and (-x));13   end;14 15 procedure swap(var a,b:longint);16   var c:longint;17   begin18     c:=a;19     a:=b;20     b:=c;21   end;22 23 procedure add(x,w:longint);24   begin25     while x<=n do26     begin27       c[x]:=max(c[x],w);28       x:=x+lowbit(x);29     end;30   end;31 32 function ask(x:longint):longint;33   begin34     ask:=0;35     while x>0 do36     begin37       ask:=max(ask,c[x]);38       x:=x-lowbit(x);39     end;40   end;41 42 procedure sort(l,r:longint);43   var y,i,j,x:longint;44   begin45     i:=l;46     j:=r;47     x:=a[(l+r) shr 1];48     y:=b[(l+r) shr 1];49     repeat50       while (b[i]>y) or (b[i]=y) and (a[i]
      
       b[j]) or (b[j]=y) and (x
       
        j) then53       begin54         swap(a[i],a[j]);55         swap(b[i],b[j]);56         inc(i);57         dec(j);58       end;59     until i>j;60     if l
        
         0 then continue;75     f[i]:=ask(a[i]-1)+1;76     ans:=max(f[i],ans);77     add(a[i],f[i]);78   end;79   writeln(ans);80 end.