【POJ2774】Long Long Message (SA)

最长公共子串...两个字符串连在一起，中间放一个特殊字符隔开。求出height之后，枚举height，看两个后缀是不是分布于两段字符串..如果是，这个值就可以作为答案。取最大值即可。

 const maxn=;
 var
  c,h,rank,sa,x,y:array[..maxn] of longint;
  n,k: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:=;
  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;
  for i:=  to n do sa[rank[i]]:=i;
 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;
 end;

 procedure init;
 var i,tot:longint;
  s1:ansistring;
 begin
  readln(s);
  s:=s+'#';
  k:=length(s);
  readln(s1);
  s:=s+s1;
  fillchar(c,sizeof(c),);
  n:=length(s);
  for i:=  to n do x[i]:=ord(s[i]);
  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;
 end;

 function range(x,y:longint):boolean;
 begin
  if (x<k) and (y>k) then exit(true);
  if (x>k) and (y<k) then exit(true);
  exit(false);
 end;

 procedure solve;
 var ans,i:longint;
 begin
  ans:=;
  for i:=  to n do if range(sa[i],sa[i-]) then ans:=max(h[i],ans);
  writeln(ans);
 end;

 Begin
  init;
  solve;
 End.