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
     
      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 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.