【BZOJ1031】字符加密Cipher（后缀数组）

题意：将一个长度为2n（复制粘贴后）的字符串的所有长度为n的后缀从小到大排序，并依次输出它们的最后一个字母。

n<=100000

思路：裸SA，模板真难背

P党不得不写成C++风格

 var ch:array[..]of char;
     a,x,y,wc,wd,sa:array[..]of longint;
     n,i,m:longint;
     tmp:char;

 function cmp(a,b,l:longint):boolean;
 begin
  exit((y[a]=y[b])and(y[a+l]=y[b+l]));
 end;

 procedure swap(var x,y:longint);
 var t:longint;
 begin
  t:=x; x:=y; y:=t;
 end;

 procedure getsa;
 var i,j,p:longint;
 begin
  for i:= to n- do
  begin
   x[i]:=a[i];
   inc(wc[a[i]]);
  end;
  for i:= to m do wc[i]:=wc[i-]+wc[i];
  for i:=n- downto  do
  begin
   dec(wc[x[i]]);
   sa[wc[x[i]]]:=i;
  end;

  j:=; p:=;
  while p<n do
  begin
   p:=;
   for i:=n-j to n- do
   begin
    y[p]:=i; inc(p);
   end;
   for i:= to n- do
    if sa[i]>=j then begin y[p]:=sa[i]-j; inc(p); end;
  // for i:= to n- do writeln(y[i]);
   for i:= to n- do wd[i]:=x[y[i]];
   for i:= to m do wc[i]:=;
   for i:= to n- do inc(wc[wd[i]]);
   for i:= to m do wc[i]:=wc[i-]+wc[i];
   for i:=n- downto  do
   begin
    dec(wc[wd[i]]);
    sa[wc[wd[i]]]:=y[i];
   end;

   for i:= to n do swap(x[i],y[i]);

   p:=; x[sa[]]:=;
   for i:= to n- do
   begin
    if cmp(sa[i-],sa[i],j) then x[sa[i]]:=p-
     else begin x[sa[i]]:=p; inc(p); end;
   end;

   j:=j*;
   m:=p;
  end;
 end;

 begin
   assign(input,'bzoj1031r.in'); reset(input);
   assign(output,'bzoj1031r.out'); rewrite(output);
  while not eoln do
  begin
   read(tmp);
   inc(n);
   ch[n]:=tmp;
  end;
  for i:= to n do a[i-]:=ord(ch[i]);
  for i:=n to n*- do a[i]:=a[i-n];
  n:=n*+; m:=;
  getsa;
  for i:= to n- do
   if sa[i]<(n>>) then
   begin
    if sa[i]= then write(chr(a[n-]))
     else write(chr(a[sa[i]-]));
   end;
  close(input);
 close(output);

 end.