红楼梦有关晴雯的情节:pascal题 迷宫

const
Road:array[1..8,1..8]of 0..3=((1,0,0,0,0,0,0,0),
(0,1,1,1,1,0,1,0),
(0,0,0,0,1,0,1,0),
(0,1,0,0,0,0,1,0),
(0,1,0,1,1,0,1,0),
(0,1,0,0,0,0,1,1),
(0,1,0,0,1,0,0,0),
(0,1,1,1,1,1,1,0)); {迷宫数组}

FangXiang:array[1..4,1..2]of -1..1=((1,0),(0,1),(-1,0),(0,-1));{四个移动方向}
WayIn:array[1..2]of byte=(1,1); {入口坐标}
WayOut:array[1..2]of byte=(8,8); {出口坐标}
Var i,j,Total:integer;

Procedure Output;
var i,j:integer;
Begin
For i:=1 to 8 do begin
for j:=1 to 8 do begin
if Road[i,j]=1 then write(#219); {1:墙}
if Road[i,j]=2 then write(' '); {2:曾走过但不通的路}
if Road[i,j]=3 then write(#03) ; {3:沿途走过的畅通的路}
if Road[i,j]=0 then write(' ') ; {0:原本就可行的路}
end; writeln;
end; inc(total); {统计总数} readln;
end;

Function Ok(x,y,i:byte):boolean; {判断坐标(X,Y)在第I个方向上是否可行}
Var NewX,NewY:shortint;
Begin
Ok:=True;
Newx:=x+FangXiang[i,1];
Newy:=y+FangXiang[i,2];
If not((NewX in [1..8]) and (NewY in [1..8])) then Ok:=False; {超界?}
If Road[NewX,NewY]=3 then ok:=false; {是否已走过的路?}
If Road[NewX,NewY]=1 then ok:=false; {是否墙?}
End;

Procedure Howgo(x,y:integer);
Var i,NewX,NewY:integer;
Begin
For i:=1 to 4 do Begin {每一步均有4个方向可选择}
If Ok(x,y,i) then Begin {判断某一方向是否可前进}
Newx:=x+FangXiang[i,1]; {前进,产生新的坐标}
Newy:=y+FangXiang[i,2];
Road[Newx,Newy]:=3; {来到新位置后,设置已走过标志}
If (NewX=WayOut[1]) and(NewY=WayOut[2]) Then Output
Else Howgo(Newx,NewY); {如到出口则输出,否则下一步递归}
Road[Newx,Newy]:=2; {堵死某一方向,不让再走,以免打转}
end;
end;
End;

Begin
total:=0;
Road[wayin[1],wayin[2]]:=3; {入口坐标设置已走标志}
Howgo(wayin[1],wayin[2]); {从入口处开始搜索}
writeln('Total is ',total); {统计总数}
end.

在说么斯？