【BZOJ4650&UOJ219】优秀的拆分（二分，hash）

题意：

思路：

 

 

在实现时SA可以用hash+二分代替，会多一个log

BZ上跑的飞快，但UOJ上extra卡出翔，已经放弃

不过转C或者写SA没准就过了

看来转C迫在眉睫

const mo=700103;
var f,g:array[0..100000]of int64;
    h,mi:array[0..100000]of int64;
    v,cas,i,n,j,x,y:longint;
    ans:int64;
    ch:ansistring;
 
function min(x,y:longint):longint;
begin
 if x<y then exit(x);
 exit(y);
end;
 
function hash(x,y:longint):longint;
begin
 hash:=(h[y]-h[x-1]*mi[y-x+1] mod mo+mo) mod mo;
end;
 
function lcp(x,y:longint):longint;
var l,r,mid,last:longint;
begin
 l:=1; r:=min(i,min(n-x+1,n-y+1)); last:=1;
 while l<=r do
 begin
  mid:=(l+r)>>1;
  if hash(x,x+mid-1)=hash(y,y+mid-1) then begin last:=mid; l:=mid+1; end
   else r:=mid-1;
 end;
 exit(last);
end;
 
function lcs(x,y:longint):longint;
var l,r,mid,last:longint;
begin
 l:=1; r:=min(i,min(x,y)); last:=1;
 while l<=r do
 begin
  mid:=(l+r)>>1;
  if hash(x-mid+1,x)=hash(y-mid+1,y) then begin last:=mid; l:=mid+1; end
   else r:=mid-1;
 end;
 exit(last);
end;
 
begin
 
 mi[0]:=1;
 for i:=1 to 50000 do mi[i]:=mi[i-1]*6662333 mod mo;
 readln(cas);
 for v:=1 to cas do
 begin
  readln(ch);
  n:=length(ch);
  for i:=0 to n+1 do
  begin
   f[i]:=0; g[i]:=0; h[i]:=0;
  end;
  for i:=1 to n do h[i]:=(h[i-1]*6662333+ord(ch[i])-ord('a')+1) mod mo;
  for i:=1 to (n+1) div 2 do
  begin
   j:=1;
   while j<=n do
   begin
    if j+i>n then break;
    if (ch[j]<>ch[j+i]) then begin j:=j+i; continue; end;
    x:=lcp(j,j+i); y:=lcs(j,j+i);
   // x:=min(x,i); y:=min(y,i);
    if x+y>i then
    begin
     inc(g[j-y+1]); dec(g[j+x-i+1]);
     inc(f[j+i-y+i]); dec(f[j+i+x]);
    end;
    j:=j+i;
   end;
  end;
  ans:=0;
  for i:=1 to n do f[i]:=f[i-1]+f[i];
  for i:=1 to n do g[i]:=g[i-1]+g[i];
  for i:=1 to n-1 do ans:=ans+f[i]*g[i+1];
  writeln(ans);
 end;
 
end.