【POJ3693】Maximum repetition substring (SA)

   这是一道神奇的题目..论文里面说得不清楚，其实是这样...如果一个长度为l的串重复多次，那么至少s[1],s[l+1],s[2*l+1],..之中有相邻2个相等...设这时为j=i*l+1,k=j+l,我们这时候借助SA和RMQ  O(1)求出：m=lcp(j,k),这时候，重复次数至少ans=m div l+1 。  当然，我们枚举到不一定能够是最优啊，因为你枚举的不一定是字符串的首尾..那这时候怎么办？就是论文里面说的，向前和向后匹配。我们设t=l-m mod l..可以理解为，这时候 m mod l为多出来的字符，t就看成是前面少的字符个数..当然如果m mod l=0就不用管这个了...那也就是说，我们再判断是否lcp(j-t,k-t)>=l   如果成立，那么 ans++ ...因为可以多出一段..   

   上面就解决了求最长的问题，下面将关于字典序的这个...其实SA就是字典序了，只要枚举是否 lcp(sa[i],sa[i]+l)>=(ans-1)*l，若成立，显然当前sa[i]起始长度为l的字符串就是答案...

   嗯..写完顿时觉得涨姿势了..

   

const maxn=100419;
var
 rec,c,h,rank,sa,x,y:array[0..maxn] of longint;
 f:array[0..maxn,0..20] of longint;
 n,cas:longint;
 s:ansistring;

function max(x,y:longint):longint; begin if x>y then exit(x) else exit(y); end;
function min(x,y:longint):longint; begin if x<y then exit(x) else exit(y); end;
procedure swap(var x,y:longint); var tmp:longint; begin tmp:=x;x:=y;y:=tmp; end;

procedure make;
var p,i,tot:longint;
begin
 p:=1;
 while p<n do
  begin
   fillchar(c,sizeof(c),0);
   for i:= 1 to n-p do y[i]:=rank[i+p];
   for i:= n-p+1 to n do y[i]:=0;
   for i:= 1 to n do inc(c[y[i]]);
   for i:= 1 to n do inc(c[i],c[i-1]);
   for i:= 1 to n do
    begin
     sa[c[y[i]]]:=i;
     dec(c[y[i]]);
    end;
   fillchar(c,sizeof(c),0);
   for i:= 1 to n do x[i]:=rank[i];
   for i:= 1 to n do inc(c[x[i]]);
   for i:= 1 to n do inc(c[i],c[i-1]);
   for i:= n downto 1 do
    begin
     y[sa[i]]:=c[x[sa[i]]];
     dec(c[x[sa[i]]]);
    end;
   for i:= 1 to n do sa[y[i]]:=i;
   tot:=1;
   rank[sa[1]]:=1;
   for i:= 2 to n do
    begin
     if (x[sa[i]]<>x[sa[i-1]]) or (x[sa[i]+p]<>x[sa[i-1]+p]) then inc(tot);
     rank[sa[i]]:=tot;
    end;
   p:=p<<1;
  end;
 for i:= 1 to n do sa[rank[i]]:=i;
end;

procedure makeh;
var i,j,p:longint;
begin
 h[1]:=0; p:=0;
 for i:= 1 to n do
  begin
   p:=max(p-1,0);
   if rank[i]=1 then continue;
   j:=sa[rank[i]-1];
   while (i+p<=n) and (j+p<=n) and (s[i+p]=s[j+p]) do inc(p);
   h[rank[i]]:=p;
  end;
end;

procedure rmq;
var i,j:longint;
begin
 for i:= 1 to n do f[i,0]:=h[i];
 for i:= 1 to trunc(ln(n)/ln(2)) do
  for j:= 1 to n-1<<i+1 do
   f[j,i]:=min(f[j,i-1],f[j+1<<(i-1),i-1]);
end;

procedure init;
var i,tot:longint;
begin
 n:=length(s);
 for i:= 1 to n do x[i]:=ord(s[i]);
 fillchar(c,sizeof(c),0);
 for i:= 1 to n do inc(c[x[i]]);
 for i:= 1 to 128 do inc(c[i],c[i-1]);
 for i:= 1 to n do
  begin
   sa[c[x[i]]]:=i;
   dec(c[x[i]]);
  end;
 rank[sa[1]]:=1;
 tot:=1;
 for i:= 2 to n do
  begin
   if x[sa[i]]<>x[sa[i-1]] then inc(tot);
   rank[sa[i]]:=tot;
  end;
 make;
 makeh;
 rmq;
end;

function lcp(x,y:longint):longint;
var t:longint;
begin
 x:=rank[x]; y:=rank[y];
 if x>y then swap(x,y);
 if x<y then inc(x);
 t:=trunc(ln(y-x+1)/ln(2));
 exit(min(f[x,t],f[y-1<<t+1,t]));
end;

procedure solve;
var m,l,i,j,tmp,t,ans,cnt:longint;
 pd:boolean;
begin
 init;
 fillchar(rec,sizeof(rec),0);
 ans:=0;
 for l:= 1 to n-1 do
  begin
   i:=1;
   while i+l<=n do
    begin
     m:=lcp(i,i+l);
     tmp:=m div l+1;
     t:=l-m mod l;
     t:=i-t;
     if (t>0) and (m mod l<>0) and (lcp(t,t+l)>=m) then inc(tmp);
     if tmp>ans then
      begin
       cnt:=1;
       rec[1]:=l;
       ans:=tmp;
      end;
     if cnt=ans then
      begin
       inc(cnt);
       rec[cnt]:=l;
      end;
     i:=i+l;
    end;
  end;
 pd:=false;
 for i:= 1 to n do
  if not pd then
   for j:= 1 to cnt do
    begin
     l:=rec[j];
     if lcp(sa[i],sa[i]+l)>=(ans-1)*l then
      begin
       t:=sa[i];
       l:=l*ans;
       pd:=true;
       break;
      end;
    end
  else break;
 inc(cas);
 write('Case ',cas,': ');
 for i:= t to t+l-1 do write(s[i]);
 writeln;
end;

Begin
 cas:=0;
 readln(s);
 while s[1]<>'#' do
  begin
   solve;
   readln(s);
  end;
End.