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
View Code
1 //Flip Game 2 const 3 win1=65535; 4 inf='1.txt'; 5 len=win1+100; 6 var 7 ac: array[0..5, 0..5]of longint=( 8 (-1,-1,-1,-1,-1,-1), 9 (-1, 1, 2, 3, 4, -1), 10 (-1, 5, 6, 7, 8, -1), 11 (-1, 9, 10, 11, 12, -1), 12 (-1, 13, 14, 15, 16, -1), 13 (-1, -1, -1, -1, -1, -1) 14 ); 15 chaos: longint; 16 ans: longint; 17 q, step: array[0..len]of longint; 18 visit: array[0..win1]of boolean; 19 procedure init; 20 var 21 i, j: longint; 22 c: char; 23 begin 24 fillchar(visit, sizeof(visit), 0); 25 chaos := 0; 26 for i := 1 to 4 do begin 27 for j := 1 to 4 do begin 28 read(c); 29 if c='b' then chaos := chaos or (1 << (ac[i, j]-1)); 30 end; 31 readln; 32 end; 33 end; 34 35 procedure main; 36 var 37 head, tail, u, i, j, tmp: longint; 38 begin 39 head := 0; tail := 1; 40 q[tail] := chaos; 41 step[tail] := 1; 42 visit[chaos] := true; 43 ans := -1; 44 if (chaos=0)or(chaos=win1) then begin ans := 0; exit; end; 45 repeat 46 inc(head); if head=len then head := 1; 47 u := q[head]; 48 for i := 1 to 4 do 49 for j := 1 to 4 do begin 50 //tmp 51 tmp := u; 52 if i>1 then tmp := tmp xor (1 << (ac[i-1, j]-1)); 53 if j<4 then tmp := tmp xor (1 << (ac[i, j+1]-1)); 54 if i<4 then tmp := tmp xor (1 << (ac[i+1, j]-1)); 55 if j>1 then tmp := tmp xor (1 << (ac[i, j-1]-1)); 56 tmp := tmp xor (1<<(ac[i, j]-1)); 57 //tmp 58 59 if visit[tmp]=false then begin 60 if (tmp=win1)or(tmp=0) then begin 61 ans := step[head]; 62 exit; 63 end; 64 visit[tmp] := true; 65 inc(tail); if tail=len then tail := 1; 66 step[tail] := step[head]+1; 67 q[tail] := tmp; 68 end; 69 end; 70 until head=tail; 71 end; 72 73 procedure print; 74 begin 75 if ans>-1 then writeln(ans) 76 else writeln('Impossible'); 77 end; 78 79 begin 80 assign(input,inf); reset(input); 81 init; 82 main; 83 print; 84 end.
