【SPOJ220】Relevant Phrases of Annihilation (SA)

成功完成3连T！   嗯没错，三道TLE简直爽到不行，于是滚去看是不是模版出问题了..拿了3份其他P党的模版扔上去，嗯继续TLE...蒟蒻表示无能为力了...

思路像论文里面说的，依旧二分长度然后分组...然后记录下每个字符的最大和最小值去判断是否满足全部成立...完事...写起来其实蛮简单的...

const maxn=100419;
var
 h,sum,rank,x,y,sa,c,lx,rx,col:array[0..maxn] of longint;
 n,k,maxlen,t,q: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,j,tot:longint;
begin
 while p<n do
  begin
   fillchar(c,sizeof(c),0);
   for i:= 1 to n-p do y[i]:=rank[i+p];
   for i:= n-p+1 to n do y[i]:=0;
   for i:= 1 to n do inc(c[y[i]]);
   for i:= 1 to n do inc(c[i],c[i-1]);
   for i:= 1 to n do
    begin
     sa[c[y[i]]]:=i;
     dec(c[y[i]]);
    end;
   fillchar(c,sizeof(c),0);
   for i:= 1 to n do x[i]:=rank[i];
   for i:= 1 to n do inc(c[x[i]]);
   for i:= 1 to n do inc(c[i],c[i-1]);
   for i:= n downto 1 do
    begin
     y[sa[i]]:=c[x[sa[i]]];
     dec(c[x[sa[i]]]);
    end;
   for i:= 1 to n do sa[y[i]]:=i;
   tot:=1;
   rank[sa[1]]:=1;
   for i:= 2 to n do
    begin
     if (x[sa[i]]<>x[sa[i-1]]) or (x[sa[i]+p]<>x[sa[i-1]+p]) then inc(tot);
     rank[sa[i]]:=tot;
    end;
   if tot=n then break;
   p:=p<<1;
  end;
 for i:= 1 to n do sa[rank[i]]:=i;
 h[1]:=0; p:=0;
 for i:= 1 to n do
  begin
   p:=max(p-1,0);
   if rank[i]=1 then continue;
   j:=sa[rank[i]-1];
   while (j+p<=n) and (i+p<=n) and (s[i+p]=s[j+p]) do inc(p);
   h[rank[i]]:=p;
  end;
end;

procedure init;
var i,j,tot,p,m:longint;
 s1:ansistring;
begin
 readln(k);
 readln(s);
 for i:= 1 to length(s) do col[i]:=1;
 maxlen:=length(s);
 for i:= 2 to k do
  begin
   readln(s1);
   maxlen:=max(length(s1),maxlen);
   s:=s+'#';
   for j:= length(s)+1 to length(s)+length(s1) do col[j]:=i;
   s:=s+s1;
  end;
 n:=length(s);
 fillchar(c,sizeof(c),0);
 for i:= 1 to n do x[i]:=ord(s[i]);
 for i:= 1 to n do inc(c[x[i]]);
 for i:= 1 to 128 do inc(c[i],c[i-1]);
 for i:= 1 to n do
  begin
   sa[c[x[i]]]:=i;
   dec(c[x[i]]);
  end;
 tot:=1;
 rank[sa[1]]:=1;
 for i:= 2 to n do
  begin
   if x[sa[i]]<>x[sa[i-1]] then inc(tot);
   rank[sa[i]]:=tot;
  end;
 make;
end;

function check(len:longint):boolean;
var i,j,t,cnt:longint;
begin
 for i:= 1 to n do
  begin
   if h[i]<len then
    begin
     fillchar(lx,sizeof(lx),$7f);
     fillchar(rx,sizeof(rx),0);
     lx[col[sa[i]]]:=sa[i];
     rx[col[sa[i]]]:=sa[i];
    end
   else
    begin
     t:=col[sa[i]];
     lx[t]:=min(lx[t],sa[i]);
     rx[t]:=max(rx[t],sa[i]);
     cnt:=0;
     for j:= 1 to k do if rx[j]-lx[j]+1>=len then inc(cnt);
     if cnt=k then exit(true);
    end;
  end;
 exit(false);
end;

procedure solve;
var l,r,mid,ans:longint;
begin
 l:=1; r:=maxlen; ans:=0;
 while l<=r do
  begin
   mid:=(l+r)>>1;
   if check(mid) then
    begin
     ans:=mid;
     l:=mid+1;
    end
   else r:=mid-1;
  end;
 writeln(ans);
end;

Begin
 readln(t);
 for q:= 1 to t do
  begin
   init;
   solve;
  end;
End.