同一个世界（三生梦醒）

前段时间写了同一个世界的，链接如下

同一个世界（erlang解题答案）

这次带来了三生梦醒的解题答案和思路，

 

首先看题目

 

还是用1代表黑色，0代表白色，用2代表黄点，x代表空白

 

抽象就是这样

----------------
| 1 | 1 | 1 | 1 |
----------------
| 1 | 0 | 0 | 1 |
----------------
| 0 | 1 | 1 | 1 |
----------------
| 1 | 0 | 0 | 0 |
----------------
| 1 | 1 | 1 | 1 |
----------------

 

这次因为每个点走的步数定了，所以暴力的使用轮询来解

把起点能走的所有路径找出来，（注意:这次的路径可以重复~~）

-module(oneword_three).

%..........省略

start(Data, {S, Scnt}, {E, Ecnt}) ->
    Pid = self(),
    Load_data = [{Num, List} || {Num, _V, List} <- Data],
    Val_data = [{Num1, V1} || {Num1, V1, _List1} <- Data, V1 =/= []],
    Tab1 = ets:new(?TAB, [public, duplicate_bag, {write_concurrency, true}]),
    Tab2 = ets:new(?TAB, [public, duplicate_bag, {write_concurrency, true}]),
    spawn(fun() -> start_try(Pid, Load_data, Tab1, S, Scnt) end),　　　　　　　　%并发去尝试所有的路径
    spawn(fun() -> start_try(Pid, Load_data, Tab2, E, Ecnt) end),
    do_loop(Val_data, Tab1, Tab2, 2).

%..........省略

start_try2(Pid, _Load_data, Tab, Loads, Nownum, 0) ->
    ets:insert(Tab, {?SIGNAL, [Nownum|Loads]}),　　　　　　%走完了就把路径存表里面
    Pid ! stop;

start_try2(Pid, Load_data, Tab, Loads, Nownum, Scnt) ->
%    ets:insert(Tab, {?SIGNAL, [Nownum|Loads]}),
    {Nownum, List} = lists:keyfind(Nownum, 1, Load_data),
    Pids = [spawn(fun() -> start_try2(Pid, Load_data, Tab, [Nownum|Loads], X, Scnt-1) end) || X <- List],
    Pid ! {start, erlang:length(Pids)},
    Pid ! stop.

 

然后就是取表里面的路径轮询，

%..........省略

start(Val_data, Tab1, Tab2) ->
    Pid = self(),
    Yellow_num = [Num || {Num, V} <- Val_data, V == 2],
    Slist = [ Z || {_N, Z}  <- ets:tab2list(Tab1)],
    Elist = [ Z1 || {_N1, Z1}  <- ets:tab2list(Tab2)],
%    io:format("here is Slist:~p, Elist:~p~n", [Slist, Elist]),
    _P = [spawn_link(fun() -> check(Pid, Yellow_num, Val_data, X, Y) end)|| X <- Slist,  Y <- Elist],　　　　%轮询去所有的数据
    wait_loop().

wait_loop() ->
    receive
        Any ->
            io:format("this is a erorr msg:~p~n", [Any])
    end,
    wait_loop().

check(Pid, Yellow_num, Val_data, X, Y) ->
    X1 = lists:reverse(X),
    Y1 = lists:reverse(Y),
    L = X1 ++ Y1,
    case check_all(Yellow_num, L) of　　　　　　　　　　　　%首先要求所有的黄点都走到
        ok ->
            case check_ok(Val_data, X1, Y1) of　　　　　　%check结果是否是全1或者全0
                ok ->
                    ?SERVERNAME ! {ok, Pid, X1, Y1};
                no ->
                    no
            end;
        no ->
            no
    end.

%..........省略

 

然后就出结果了

13> oneword:start(A,{12,5},{13,5}).               
<0.6761.106>
 ----------------
| 1 | 1 | 1 | 1 |
 ----------------
| 1 | 0 | 0 | 1 |
 ----------------
| 0 | 1 | 1 | 1 |
 ----------------
| 1 | 0 | 0 | 0 |
 ----------------
| 1 | 1 | 1 | 1 |
 ----------------
the answere is:
 line1:[12,11,10,6,7,11]
 line2:[13,9,10,14,15,16]

 

tips：

如果有3个起点，可以先随便走一个，猜其他的2个，如下

 

走一下下面步骤3的，变成这样

然后再解答

16> oneword:start(A,{5,5},{8,5}).  
<0.11334.106>
 ----------------
| X | 1 | 1 | X |
 ----------------
| 1 | 0 | 0 | 1 |
 ----------------
| 1 | 0 | 0 | 1 |
 ----------------
| 1 | 1 | 1 | 1 |
 ----------------
| X | 0 | 0 | X |
 ----------------
the answere is:
 line1:[5,6,10,14,18,14]
 line2:[8,7,11,15,19,15]

答案就出来了