【SPOJ694】Distinct Substrings （SA）

求不相同子串个数    该问题等价于求所有后缀间不相同前缀的个数..也就是对于每个后缀suffix(sa[i])，将贡献出n-sa[i]+1个，但同时，要减去那些重复的，即为height[i]，故答案为n-sa[i]+1-height[i]的累计。

const maxn=;

var
 x,y,rank,sa,h,c:array[..maxn] of longint;
 s:ansistring;
 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 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 makeht;
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,j,tot:longint;
 ch:char;
begin
 readln(s);
 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;
 makeht;
end;

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

Begin
 readln(t);
 for q:=  to t do
  begin
   init;
   solve;
  end;
End.