ZJU2112 Dynamic Rank

Dynamic Rank

線段樹套平衡樹～

先二分答案，找到g[l]<kth<=g[r],（其中g[x]表示在全部的線段的平衡樹中小於等於x的數的個數）

則r爲答案～要注意二分邊界和終止條件～

其中的平衡樹是用sbt退化版寫的，線段樹是用二叉堆的方法 存的～

 var seg,t:array [1..1000000] of longint; left,right,size,data:array [0..1000000] of longint; tmp,tot,tot2,n,m,i,tt:longint; procedure updata(t:longint); begin if t<>0 then size[t]:=size[left[t]]+size[right[t]]+1; end; procedure leftR(var t:longint); begin tmp:=left[t]; left[t]:=right[tmp]; right[tmp]:=t; updata(t); updata(tmp); t:=tmp; end; procedure rightR(var t:longint); begin tmp:=right[t]; right[t]:=left[tmp]; left[tmp]:=t; updata(t); updata(tmp); t:=tmp; end; procedure insert(var t:longint;v:longint); begin if t=0 then begin inc(tot); left[tot]:=0; right[tot]:=0; size[tot]:=1; data[tot]:=v; t:=tot; end else begin inc(size[t]); if data[t]>v then begin insert(left[t],v); if size[left[left[t]]]>size[right[t]] then leftR(t); end else begin insert(right[t],v); if size[right[right[t]]]>size[left[t]] then rightR(t); end; end; end; function delete(var t:longint;v:longint):longint; begin dec(size[t]); if (data[t]=v) or (data[t]>v) and (left[t]=0) or (data[t]<v) and(right[t]=0) then begin delete:=data[t]; if (left[t]=0) or (right[t]=0) then t:=left[t]+right[t] else data[t]:=delete(left[t],data[t]+1); end else begin if data[t]>v then delete:=delete(left[t],v) else delete:=delete(right[t],v); end; end; procedure insseg(tt,l,r,k,v:longint); begin if l=r then begin insert(seg[tt],v); end else begin if k>((l+r)>>1) then begin insseg(tt<<1+1,(l+r)>>1+1,r,k,v); insert(seg[tt],v); end else begin insseg(tt<<1,l,(l+r)>>1,k,v); insert(seg[tt],v); end; end; end; procedure init; var i,j:longint; begin fillchar(seg,sizeof(seg),0); readln(n,m); tot2:=0; tot:=0; for i:=1 to n do begin read(j); insseg(1,1,n,i,j); end; readln; end; function getmin(t:longint):longint; begin while left[t]<>0 do t:=left[t]; getmin:=data[t]; end; function getmax(t:longint):longint; begin while right[t]<>0 do t:=right[t]; getmax:=data[t]; end; procedure getseg(tt,l,r,ll,rr:longint); begin if (ll>r) or(rr<l) then exit; if (ll<=l) and (r<=rr) then begin inc(tot2); t[tot2]:=seg[tt]; end else begin getseg(tt<<1,l,(l+r)>>1,ll,rr); getseg(tt<<1+1,(l+r)>>1+1,r,ll,rr); end; end; function change(tt,l,r,k,v:longint):longint; begin if l=r then begin change:=data[seg[tt]]; data[seg[tt]]:=v; end else begin if k>((l+r)>>1) then begin change:=change(tt<<1+1,(l+r)>>1+1,r,k,v); delete(seg[tt],change); insert(seg[tt],v); end else begin change:=change(tt<<1,l,(l+r)>>1,k,v); delete(seg[tt],change); insert(seg[tt],v); end; end; end; function get(t:longint;v:longint):longint; begin if t=0 then exit(0) else if data[t]>v then get:=get(left[t],v) else get:=size[left[t]]+1+get(right[t],v); end; function succ(t,v:longint):longint; begin if t=0 then exit(v); if data[t]<v then succ:=succ(right[t],v) else begin succ:=succ(left[t],v); if succ=v then succ:=data[t]; end; end; procedure doit(a,b,kth:longint); var i,j,l,r,mid,ans:longint; begin tot2:=0; getseg(1,1,n,a,b); l:=maxlongint; r:=-1; for i:=1 to tot2 do begin j:=getmin(t[i]); if j<l then l:=j; j:=getmax(t[i]); if j>r then r:=j; end; dec(l); inc(r); // l:=-1; // r:=maxlongint; repeat mid:=(l+r) div 2; j:=0; for i:=1 to tot2 do inc(j,get(t[i],mid)); if j>=kth then r:=mid else l:=mid; until l+1=r; {ans:=maxlongint; for i:=1 to tot2 do begin j:=succ(t[i],r); if j<ans then ans:=j; end; writeln(ans);} writeln(r); end; procedure work; var i,j,k,l:longint; ch:char; begin for l:=1 to m do begin read(ch); if ch='C' then begin readln(i,j); change(1,1,n,i,j); end else begin readln(i,j,k); doit(i,j,k); end; end; end; begin assign(input,'in.in'); assign(output,'out.out'); reset(input); rewrite(output); readln(tt); for i:=1 to tt do begin init; work; end; close(input); close(output); end.

這道題寫了很久～其實就是二分的問題～二分正確了其他的只是時間問題～