Poj 1753

题目大意：翻转棋游戏，在一个4*4的棋盘上，每个格子有两个颜色黑或白，把该个格子翻转（黑变白白变黑），同时它的上下左右4个格子也会如此，求把整个棋盘的颜色统一的方案，没有输出impossible

解：状态压缩bfs，弱死了，首先是标记二掉了，一种状态出现过就不用取消了，二是忘记输出无解，wa+tle..,弱弱弱弱弱弱弱弱弱

//Flip Game
const
        win1=65535;
        inf='1.txt';
        len=win1+100;
var
        ac: array[0..5, 0..5]of longint=(
                                        (-1,-1,-1,-1,-1,-1),
                                        (-1, 1, 2, 3, 4, -1),
                                        (-1, 5, 6, 7, 8, -1),
                                        (-1, 9, 10, 11, 12, -1),
                                        (-1, 13, 14, 15, 16, -1),
                                        (-1, -1, -1, -1, -1, -1)
                                        );
        chaos: longint;
        ans: longint;
        q, step: array[0..len]of longint;
        visit: array[0..win1]of boolean;
procedure init;
var
        i, j: longint;
        c: char;
begin
  fillchar(visit, sizeof(visit), 0);
  chaos := 0;
  for i := 1 to 4 do begin
    for j := 1 to 4 do begin
      read(c);
      if c='b' then chaos := chaos or (1 << (ac[i, j]-1));
    end;
    readln;
  end;
end;

procedure main;
var
        head, tail, u, i, j, tmp: longint;
begin
  head := 0; tail := 1;
  q[tail] := chaos;
  step[tail] := 1;
  visit[chaos] := true;
  ans := -1;
  if (chaos=0)or(chaos=win1) then begin ans := 0; exit; end;
  repeat
    inc(head); if head=len then head := 1;
    u := q[head];
    for i := 1 to 4 do
      for j := 1 to 4 do begin
        //tmp
        tmp := u;
        if i>1 then tmp := tmp xor (1 << (ac[i-1, j]-1));
        if j<4 then tmp := tmp xor (1 << (ac[i, j+1]-1));
        if i<4 then tmp := tmp xor (1 << (ac[i+1, j]-1));
        if j>1 then tmp := tmp xor (1 << (ac[i, j-1]-1));
        tmp := tmp xor (1<<(ac[i, j]-1));
        //tmp

        if visit[tmp]=false then begin
          if (tmp=win1)or(tmp=0) then begin
            ans := step[head];
            exit;
          end;
          visit[tmp] := true;
          inc(tail); if tail=len then tail := 1;
          step[tail] := step[head]+1;
          q[tail] := tmp;
        end;
      end;
  until head=tail;
end;

procedure print;
begin
  if ans>-1 then writeln(ans)
    else writeln('Impossible');
end;

begin
  assign(input,inf); reset(input);
  init;
  main;
  print;
end.