【SPOJ1297】Palindrome (SA+RMQ)

求最长回文串。把原串翻转后，加在原串后面，中间插入一个辨别字符。然后求SA,Height。然后枚举每个字母作为回文串中心，分长度为奇数和偶数去讨论:奇数求 suffix(i)和suffix(n-i+1)的最长公共前缀，偶数则求suffix(i)和suffix(n-i+2)(当然，i=1时不成立) 。然后问题就是求最长公共前缀了，转换为RMQ问题，O(nlogn)预处理，O(1)询问即可.

 const maxn=;

 var
  x,y,rank,sa,h,c:array[..maxn] of longint;
  s:ansistring;
  f:array[..maxn,..] of longint;
  t,q,n:longint;

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

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

 procedure rmq;
 var i,j:longint;
 begin
  fillchar(f,sizeof(f),$7f);
  for i:=  to n do f[i,]:=h[i];
  for i:=  to trunc(ln(n)/ln()) do
   for j:=  to n-<<i+ do
    f[j,i]:=min(f[j,i-],f[j+<<(i-),i-]);
 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 x:=x+;
  t:=trunc(ln(y-x+)/ln());
  exit(min(f[x,t],f[y-<<t+,t]));
 end;

 procedure init;
 var i,j,tot:longint;
  ch:char;
 begin
  readln(s);
  s:=s+'#';
  for i:= length(s)- downto  do s:=s+s[i];
  n:=length(s);
  for i:=  to n do x[i]:=ord(s[i]);
  fillchar(c,sizeof(c),);
  for i:=  to n do inc(c[x[i]]);
  for i:=  to  do inc(c[i],c[i-]);
  for i:=  to n do
   begin
    sa[c[x[i]]]:=i;
    dec(c[x[i]]);
   end;
  rank[sa[]]:=;
  tot:=;
  for i:=  to n do
   begin
    if x[sa[i]]<>x[sa[i-]] then inc(tot);
    rank[sa[i]]:=tot;
   end;
  make;
  makeh;
  rmq;
 end;

 procedure solve;
 var ans,st,i,k:longint;
 begin
  ans:=;st:=;
  for i:=  to n do
   begin
    k:=lcp(i,n-i+);
    if k*->ans then
     begin
      st:=i-k+;
      ans:=k*-;
     end;
    if i> then
     begin
      k:=lcp(i,n-i+);
      if k*>ans then
       begin
        st:=i-k;
        ans:=k*;
       end;
     end;
   end;
  for i:= st to st+ans- do write(s[i]);
 end;

 Begin
  init;
  solve;
 End.