让人纠结的8数码问题。。

program aa;
   var
   boo:boolean;
   i,j,k,p,u,du,px,k1:integer;
   f:text;
   xr:array[1..50] of integer;
   pr:array[1..9] of integer;
   orr:array[1..2,1..50] of integer;
   cr:array[1..9] of integer;
   br:array[1..9,1..50] of   integer;
    ar:array[1..3,1..3] of integer;
    dr:array[1..10] of integer;
    procedure swap(var no1,no2:integer);
      var
      kong:integer;
       begin
       kong:=0;
        kong:=no1;
        no1:=no2;
        no2:=kong;
       end;
         procedure huifu;
           var
           y:integer;
            begin
            y:=0;
            for i:=1 to 3 do
                 for j:=1 to 3 do
                 begin
                 inc(y);
             ar[i,j]:= br[y,k];
             end;
            end;
             procedure zhuanhua;
              var
              x:integer;
                begin
                inc(k);
                x:=0;
                for i:=1 to 3 do
                 for j:=1 to 3 do
                 begin
                  inc(x);
                  br[x,k]:=ar[i,j];


                  end;
                  end;
                 procedure same;
                 var
                 p1:integer;
                 ji:integer;
                   begin
                   p1:=0;
                   for i:=1 to 3 do
                   for j:=1 to 3 do
                    begin
                       inc(p1);
                      pr[p1]:=ar[i,j]
                    end;
                   boo:=true;
                   for j:=1 to k do
                   begin
                    for i:=1 to 9 do
                    if br[i,j]=pr[i] then
                       xr[j]:=xr[j]+1;
                     if xr[j]=9   then
                      boo:=false;
                    end;
                    end;
           procedure pan0;


             begin
               for i:=1 to 3 do
                for j:=1 to 3 do
                 if ar[i,j]=0 then
                  begin
                 orr[1,p]:=i;
                   orr[2,p]:=j;
                  end;
             end;
    begin
    k:=1;
    p:=1;
    assign(f,'sun.txt');
     reset(f);
     for i:=1 to 3 do
     begin
      for j:=1 to 3 do
      read(f,ar[i,j]);
       {if ar[i,j]=0 then begin
       orr[1,p]:=i;
       orr[2,p]:=j;
       end;}
      end;
      readln(f);
       for i:=1 to 9 do
       read(f,cr[i]);
for i:=1 to 3 do
begin
for j:=1 to 3 do
begin
inc(px);
pr[px]:=ar[i,j];
end;
end;
    for i:=1 to 9 do
    begin
    k1:=k1+1;
      br[k1,k]:=pr[k1]
      end;
     repeat
     pan0;
      if orr[1,p]>1 then begin
       swap(ar[orr[1,p],orr[2,p]],ar[orr[1,p]-1,orr[2,p]]);
       same;
        if boo=true   then begin
           zhuanhua;{zhuan hua dao br[] li bian}
          swap(ar[orr[1,p],orr[2,p]],ar[orr[1,p]-1,orr[2,p]]);
    {hui fu mei you jiao huan de er wei biao}
           end;
           end;
           if orr[1,p]<3 then begin
             swap(ar[orr[1,p],orr[2,p]],ar[orr[1,p]+1,orr[2,p]]);
             same;


             if   boo=true then begin
            zhuanhua;
              swap(ar[orr[1,p],orr[2,p]],ar[orr[1,p]+1,orr[1,p]]);
         {hui fu mei you jiao huan de er wei biao}
            end;
            end;
             if orr[2,p]>1 then begin
                swap(ar[orr[1,p],orr[2,p]],ar[orr[1,p],orr[2,p]-1]);
              same;
              if boo=true then begin
             zhuanhua;
            swap(ar[orr[1,p],orr[2,p]],ar[orr[1,p]+1,orr[2,p]-1]);
              {hui fu mei you jiao huan de er wei biao}
             end;
             end;
              if orr[2,p]<3 then   begin
             swap(ar[orr[1,p],orr[2,p]],ar[orr[1,p],orr[2,p]+1]);
                same;
                if boo=true then   begin
                zhuanhua;
                     swap(ar[orr[1,p],orr[2,p]],ar[orr[1,p],orr[2,p]+1]);
             {hui fu mei you jiao huan de er wei biao}
               end;
               end;
               inc(p);
                 for i:=1 to 3 do
                  for j:=1 to 3 do
                  begin
                  inc(u);
                 ar[i,j]:=br[u,p];
                       end;
                   dr[p]:=k;{dr li cun de shi mei ci fen zhi de shu   ji bu shi 3 jiu shi 4,bian yu da yin jie guo}
       until
   (br[1,k]=cr[1])and (br[2,k]=cr[2]) and (br[3,k]=cr[3])and( br[4,k]=cr[4])and
     (br[5,k]=cr[5])and    (br[6,k]=cr[6])and    (br[7,k]=cr[7])and
      (br[8,k]=cr[8])and   (br[9,k]=cr[9]);
assign(f,'rui.txt');
rewrite(f);
write(f,p);
            end.

回复：2楼
program aa;
    var
    boo:boolean;
    i,j,k,p,u,du,px,k1:integer;
    f:text;
    xr:array[1..50] of integer;
    pr:array[1..9] of integer;
    orr:array[1..2,1..50] of integer;
    cr:array[1..9] of integer;
    br:array[1..9,1..50] of    integer;
     ar:array[1..3,1..3] of integer;
     dr:array[1..10] of integer;
     procedure swap(var no1,no2:integer);
       var
       kong:integer;
        begin
        kong:=0;
         kong:=no1;
         no1:=no2;
         no2:=kong;
        end;
          procedure huifu;
            var
            y:integer;
             begin
             y:=0;
             for i:=1 to 3 do
                  for j:=1 to 3 do
                  begin
                  inc(y);
              ar[i,j]:= br[y,k];
              end;
             end;
              procedure zhuanhua;
               var
               x:integer;
                 begin
                 inc(k);
                 x:=0;
                 for i:=1 to 3 do
                  for j:=1 to 3 do
                  begin


                   inc(x);
                   br[x,k]:=ar[i,j];
2010-7-16 16:22 回复  
sr_tp
0位粉丝
6楼
                   end;
                   end;
                  procedure same;
                  var
                  p1:integer;
                  ji:integer;
                    begin
                    p1:=0;
                    for i:=1 to 3 do
                    for j:=1 to 3 do
                     begin
                        inc(p1);
                       pr[p1]:=ar[i,j]
                     end;
                    boo:=true;
                    for j:=1 to k do
                    begin
                     for i:=1 to 9 do
                     if br[i,j]=pr[i] then
                        xr[j]:=xr[j]+1;
                      if xr[j]=9    then


                       boo:=false;
                     end;
                     end;
            procedure pan0;
2010-7-16 16:22 回复  
sr_tp
0位粉丝
7楼
              begin
                for i:=1 to 3 do
                 for j:=1 to 3 do
                  if ar[i,j]=0 then
                   begin
                  orr[1,p]:=i;
                    orr[2,p]:=j;
                   end;
              end;
     begin
     k:=1;
     p:=1;
     assign(f,'sun.txt');
      reset(f);
      for i:=1 to 3 do
      begin
       for j:=1 to 3 do
       read(f,ar[i,j]);
        {if ar[i,j]=0 then begin
        orr[1,p]:=i;
        orr[2,p]:=j;
        end;}
       end;
       readln(f);
        for i:=1 to 9 do
        read(f,cr[i]);
for i:=1 to 3 do
begin
for j:=1 to 3 do
begin
inc(px);
pr[px]:=ar[i,j];
end;
end;
     for i:=1 to 9 do
     begin
     k1:=k1+1;
       br[k1,k]:=pr[k1]
       end;
      repeat
      pan0;
       if orr[1,p]>1 then begin
        swap(ar[orr[1,p],orr[2,p]],ar[orr[1,p]-1,orr[2,p]]);
        same;


                 zhuanhua;
                      swap(ar[orr[1,p],orr[2,p]],ar[orr[1,p],orr[2,p]+1]);
              {hui fu mei you jiao huan de er wei biao}
                end;
                end;
                inc(p);
                  for i:=1 to 3 do
                   for j:=1 to 3 do
                   begin
                   inc(u);
                  ar[i,j]:=br[u,p];
                        end;
                    dr[p]:=k;{dr li cun de shi mei ci fen zhi de shu    ji bu shi 3 jiu shi 4,bian yu da yin jie guo}
        until
    (br[1,k]=cr[1])and (br[2,k]=cr[2]) and (br[3,k]=cr[3])and( br[4,k]=cr[4])and
      (br[5,k]=cr[5])and     (br[6,k]=cr[6])and     (br[7,k]=cr[7])and
       (br[8,k]=cr[8])and    (br[9,k]=cr[9]);
assign(f,'rui.txt');
rewrite(f);
write(f,p);
             end.