同一个世界(erlang解题答案)

最近玩同一个世界,才几关就把3次提示用完了,十分气愤,

于是写了程序来解~~~

o(^▽^)o

 

以2-1为例子,题目如下

 

以0代表白色,1代表黑色,抽象的就是这样的

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

并发N个进程去走, 走一步,就判断剩下的白色或者黑色的是否能连成一条线,

 下面上程序

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%...............................<br><br>start_now(Nowdata, Begnum, Endnum) ->
    P1 = spawn(fun() ->start_one(Nowdata, Begnum, Endnum, 0) end),    %假设最后全黑或者全白
    P3 = spawn(fun() ->start_one(Nowdata, Begnum, Endnum, 1) end),
    loop([P1, P3]).
 
start_one(Nowdata, Begnum, Endnum, Flag) ->
    Pid = self(),
    start_two(Pid, Nowdata, Begnum, Endnum, Flag),
    wait_loop(Pid).
 
start_two(Pid, Nowdata, Begnum, Endnum, Flag) ->
    Lin = splite_list(Nowdata, Endnum, Flag),
    spawn(fun() -> oneword_two:start(Pid, Nowdata, Begnum, Endnum, Lin) end). %并发进程去走<br><br>%...............................

 

然后就是走和判断了

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%...............................<br>start_try(Fid, Data, Loadlist, Nownum, Endnum, Lin) ->
    if [Nownum|Loadlist] == [3,9,8,7,13,19,20,14,15,16,17,23,29,28,22,21,27,26,25,31,32,
                            33,34,35,36,42] ->
        io:format("here is tttttt Loadlist:~p, Nownum:~p~n,Data:~p~n", [Loadlist, Nownum, Data]);
        true -> ok
    end,
%   io:format("here is 2222 Loadlist:~p, Nownum:~p,Data:~p~n", [Loadlist, Nownum, Data]),
    try link(Fid) of
        true ->
            case maybe_end(Endnum, [Nownum|Loadlist], Lin) of
                false ->
                    false;
                true ->
                    {ok, Enlist, Data2} = move(Data, Loadlist, Nownum),        %%走一个位置
                    case oneword_check:check_ok(Fid, Data2, Endnum) of        %%判断剩下的是否能连成一条线
                        {right, Line} ->
                            ?SERVERNAME ! {ok, {[Nownum|Loadlist], Line}};
                        wrong ->
        %                   oneword_lib:sleep(1),
                            [spawn(fun() -> start_try(Fid, Data2, [Nownum|Loadlist], X, Endnum, Lin) end)|| X <- Enlist]  %%如果不能,再并发N个进程去走下一步~~
                    end    
            end
        catch _:_ ->
                ok
        end.<br><br>%...............................

  

然后结果就是很简单了下

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7> rp(A).
[0,1,0,1,0,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,0,1,1,1,0,0,0,0,0,
 0]
ok
8> oneword:start(A,2,15).
<0.62.0>
 ----------------------------
| 0 | 1 | 0 | 1 | 0 |
 ----------------------------
| 0 | 1 | 0 | 1 | 0 |
 ----------------------------
| 1 | 0 | 1 | 0 | 1 |
 ----------------------------
| 0 | 1 | 0 | 1 | 0 |
 ----------------------------
| 0 | 1 | 1 | 1 | 0 |
 ----------------------------
| 0 | 0 | 0 | 0 | 0 |
 ----------------------------
here is 111 Begnum:2,Endnum:15, Lin:false
here is 111 Begnum:2,Endnum:15, Lin:true
the answere is:                %答案1,
 line1:[2,3,8,9,14,13,12,11]
 line2:[15,14,19,24,23,22,17,12,7,8,3,4]
9> oneword:start(A,2,15).
<0.8545.1>
 ----------------------------
| 0 | 1 | 0 | 1 | 0 |
 ----------------------------
| 0 | 1 | 0 | 1 | 0 |
 ----------------------------
| 1 | 0 | 1 | 0 | 1 |
 ----------------------------
| 0 | 1 | 0 | 1 | 0 |
 ----------------------------
| 0 | 1 | 1 | 1 | 0 |
 ----------------------------
| 0 | 0 | 0 | 0 | 0 |
 ----------------------------
here is 111 Begnum:2,Endnum:15, Lin:false
here is 111 Begnum:2,Endnum:15, Lin:true
the answere is:                %%答案2
 line1:[2,7,12,17,22,23,24,19,14,9,4]
 line2:[15,14,13,12,11]
1
 

 1,2,3,4,5,

   6,7,8,9,10,

   11, 12,13,14, 15

 ........

答案的序号是格子的编号代表,格子的顺序,从1开始,如下

发现这种答案不是唯一的~~~~

 

1
 
posted @   土豆008  阅读(734)  评论(0编辑  收藏  举报
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