3传教士与3野人过河问题的A*算法

clc;

clear;

global State;

 

%此程序计算传教士与野人问题：

%三个传教士与三个野人分别站在河的两岸，有一条船，可以载一至两人。要求用船载人，把三个传教士、野人载过岸，要求每个地方的野人数量不能大于传教士

% 初始状态：

%    'Fa1'  '   '  'Sa1'

%    'Fa2'  '   '  'Sa2'

%    'Fa3'          'Sa3'

% 

% 最终状态

%    'Sa1'  '   '  'Fa1'

%    'Sa2'  '   '  'Fa2'

%    'Sa3'          'Fa3'

 

%定义初始状态

%河的左岸为3个传教士，0个野人

%河的右岸为0个传教士，3个野人

State_Begin =  { 'Fa' , 'Fa', 'Fa' 'Sa', 'Sa', 'Sa' 'none','none'};

 

%最终状态

%河的左岸为0个传教士，3个野人

%河的右岸为3个传教士，0个野人

%State_Final =  { 'Sa', 'Sa', 'Sa'  'Fa' , 'Fa', 'Fa' 'none', 'none'};

 

%生成State的序列，结果就是State_Begin -> State1 -> State2 -> State3 -> .....->State_Final

    %构建状态点，包含当前节点编号(No_now), 父节点编号（No_dad）,f,g,是否已被展开过（1为展开过，0为未展开）,当前状态(8个状态变量)

    %初始状态

   State ={'No_now','No_dad','f','g','Flag','State_L1','State_L2','State_L3','State_R1','State_R2','State_R3','State_B1','State_B2'};

    

   g=1;

    h= fun_h(State_Begin);

    f= h + g;

   Flag = 0;  %未被展开

   State =[State ;[1,1,f,g,Flag,State_Begin]];

    

   while h~=0

       %展开结点，并将展开的节点状态从后面加入State

       ind = find(cell2mat(State(2:end,5))==0); %找到没有被展开的接点

       %由没展开的接点组成新的State

       State_new = State(ind+1 , :);

       

       fmin = min(cell2mat(State_new(:,3)));

       

       ind_new = find(cell2mat(State_new(:,3)) == fmin);

                

       if length(ind_new)==1

           No_now_tmp = ind_new;

           else

           No_now_tmp = ind_new(1);

       end

       

       No_now = State_new{No_now_tmp,1};

 

       fun_unfold(No_now);

       %判断是否获得了终点状态

       mat_h = cell2mat(State(2:end,3)) - cell2mat(State(2:end,4));

       indh = find(mat_h == 0);

        if isempty(indh)==1

           h = 1;

       else

           h = 0;

       end

       

    end

   

 

% %显示结果

% 数据格式为{'No_now','No_dad','f','g','Flag','State_L1','State_L2','State_L3','State_R1','State_R2','State_R3','State_B1','State_B2'}

 

mat_h = cell2mat(State(2:end,3)) -cell2mat(State(2:end,4));

ind = find( mat_h == 0 );  

       if length(ind)==1

           No_now_tmp = ind;

       else

           No_now_tmp = ind(1);

       end

No_now = State{No_now_tmp+1,1};

disp(State(No_now_tmp+1,6:end));

while No_now ~= 1

   No_now = State{No_now+1,2};

   ind =  find(cell2mat(State(2:end,1)) == No_now );

   disp('                       /\  /\    ');

   disp('                       |   |    ');

   disp(State(ind+1,6:end));

end

 

function fun_unfold(No_now)

    

   global State;

    %数据格式为{'No_now','No_dad','f','g','Flag','State_L1','State_L2','State_L3','State_R1','State_R2','State_R3','State_B1','State_B2'}

   State_unfold = State(No_now+1,:);

    

    %每展开一次，层数就相对上层增加1层

    g= State_unfold{4} + 1;

    

   for i=6:11

       for j=6:11

           State_tmp = State_unfold;

          

           tmp = State_tmp{i};

           State_tmp{i} = State_tmp{12};

           State_tmp{12} = tmp;

 

           tmp = State_tmp{j};

           State_tmp{j} = State_tmp{13};

           State_tmp{13} = tmp;

 

           check = fun_check(State_tmp(6:13));

           if check == 1

                f =  fun_h(State_tmp(6:13)) + g;

                State_tmp{3} = f; 

                State_tmp{4} = g;

                State_tmp{2} = No_now ;   % 父节点即为No_now 结点

                [r,~] = size(State);

                State_tmp{1} = r;

                State = [State ; State_tmp];

           end

           

       end

   end

   State{No_now+1,5} = 1;

end

 

function [check]=fun_check(State_now)

    %分别统计两岸的野人数和神父数，要求野人数不要大于神父数，否则返回0

    %左岸检查

   Fa1=0;

   Sa1=0;

   check1 = 0;

   check2 = 0;

   for i=1:3

       if strcmp(State_now{i},'Fa')==1

 

           Fa1 = Fa1+1;

       elseif strcmp(State_now{i},'Sa')==1

           Sa1 = Sa1+1;

       end

   end

    

   if (Fa1 ~= 0) && (Sa1 >Fa1)

       check = 0;

   else

       check1 = 1;

   end

 

    %右岸检查

   Fa2=0;

   Sa2=0;

   for i=4:6

       if strcmp(State_now{i},'Fa')==1

           Fa2 = Fa2+1;

       elseif strcmp(State_now{i},'Sa')==1

           Sa2 = Sa2+1;

       end

   end

    

   if (Fa2 ~= 0) && (Sa2 >Fa2)

       check = 0;

   else

       check2 = 1;

   end

       

   if check2 == 1 && check1 ==1

       check =1;

 

   else

       check =0;

 

   end

 

end

 

function h = fun_h (state)

 %%h为从当前状态到目标状态的代价

 

 %最终状态

%河的左岸为0个传教士，3个野人

%河的右岸为3个传教士，0个野人

%State_Final =  { 'Sa', 'Sa', 'Sa'  'Fa' , 'Fa', 'Fa' 'none', 'none'};

h = 0;

for i=1:3

   if strcmp(state{i},'Sa')~=1

       h = h + 1;

   end

end

 

for i=4:6

   if strcmp(state{i},'Fa')~=1

       h = h + 1;

   end

end

end 

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