题意:
1 var f,p:array[0..1100000]of longint; 2 a:array[0..1100000]of char; 3 len,i,n,mx,id,ans,j:longint; 4 ch:ansistring; 5 6 function min(x,y:longint):longint; 7 begin 8 if x<y then exit(x); 9 exit(y); 10 end; 11 12 function max(x,y:longint):longint; 13 begin 14 if x>y then exit(x); 15 exit(y); 16 end; 17 18 function find(k:longint):longint; 19 begin 20 if f[k]<>k then f[k]:=find(f[k]); 21 find:=f[k]; 22 end; 23 24 begin 25 assign(input,'bzoj2342.in'); reset(input); 26 assign(output,'bzoj2342.out'); rewrite(output); 27 readln(len); 28 readln(ch); 29 n:=2; a[1]:='@'; a[2]:='#'; 30 for i:=1 to len do 31 begin 32 inc(n); a[n]:=ch[i]; 33 inc(n); a[n]:='#'; 34 end; 35 inc(n); a[n]:='$'; 36 mx:=0; id:=0; 37 for i:=2 to n-1 do 38 begin 39 if mx>i then p[i]:=min(p[id*2-i],mx-i) 40 else p[i]:=1; 41 while a[i-p[i]]=a[i+p[i]] do inc(p[i]); 42 if p[i]+i>mx then 43 begin 44 mx:=p[i]+i; 45 id:=i; 46 end; 47 end; 48 for i:=2 to n-1 do 49 if a[i]='#' then f[i]:=i 50 else f[i]:=i+1; 51 i:=2; j:=2; 52 repeat 53 i:=i+2; 54 if i>n then break; 55 j:=find(max(i-p[i] div 2,2)); 56 while (j<i)and(j+p[j]<i) do 57 begin 58 f[j]:=find(j+1); 59 j:=f[j]; 60 end; 61 if j<i then ans:=max(ans,(i-j)*2); 62 until i>n; 63 writeln(ans); 64 close(input); 65 close(output); 66 end.