vijos 1197 费解的开关 反向bfs+位运算

先反向bfs算出目标状态到每种状态的最少步数，然后每读入一个就直接输出就好了。

代码：

const
  maxm=33554432;
  maxn=250000;

var
  s,i,j,n,l:longint;
  c:char;
  v:array[0..maxm] of 0..6;
  state,f:array[1..maxn] of longint;

procedure bfs;
var
  head,tail,i,u,x:longint;
begin
  head:=0;
  tail:=1;
  state[1]:=1 shl 25-1;
  f[1]:=0;
  repeat
    inc(head);
    u:=state[head];
    if v[u]=6 then exit;;
    for i:=0 to 24 do
    begin
      x:=u xor (1 shl i);
      if i mod 5<>0 then x:=x xor (1 shl (i-1));
      if (i+1) mod 5<>0 then x:=x xor (1 shl (i+1));
      if i<20 then x:=x xor (1 shl (i+5));
      if i>4 then x:=x xor (1 shl (i-5));
      if v[x]=0 then
      begin
        inc(tail);
        state[tail]:=x;
        v[x]:=v[u]+1;
        if x=0 then exit;
      end;
    end;
  until head>=tail;
end;

begin
  readln(n);
  bfs;
  for l:=1 to n do
  begin
    s:=0;
    for i:=1 to 5 do
      for j:=1 to 5 do
      begin
        if j<5
          then read(c)
          else readln(c);
        if c='1' then s:=s+(1 shl ((i-1)*5+j-1));
      end;
    if l<n then readln;
    if s=1 shl 25-1
      then writeln(0)
      else if v[s]=0
             then writeln(-1)
             else writeln(v[s]);
  end;
end.