电子老鼠走迷宫

最新推荐文章于 2024-05-25 09:46:25 发布

原创最新推荐文章于 2024-05-25 09:46:25 发布 · 525 阅读

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Title 
 电子老鼠走迷宫
Description 
如下图12×12方格图，找出一条自入口（2，9）到出口（11，8）的最短路径。 
Input 
Output 
Sample Input 
12  //迷宫大小
2 9 11 8 //起点和终点
1 1 1 1 1 1 1 1 1 1 1 1  //邻接矩阵，0表示通，1表示不通
1 0 0 0 0 0 0 1 0 1 1 1
1 0 1 0 1 1 0 0 0 0 0 1
1 0 1 0 1 1 0 1 1 1 0 1
1 0 1 0 0 0 0 0 1 0 0 1
1 0 1 0 1 1 1 1 1 1 1 1
1 0 0 0 1 0 1 0 0 0 0 1
1 0 1 1 1 0 0 0 1 1 1 1
1 0 0 0 0 0 1 0 0 0 0 1
1 1 1 0 1 1 1 1 0 1 0 1
1 1 1 1 1 1 1 0 0 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1
Sample Output 
(2,9)->(3,9)->(3,8)->(3,7)->(4,7)->(5,7)->(5,6)->(5,5)->(5,4)->(6,4)->(7,4)->(7,3)->(7,2)->(8,2)->(9,2)->(9,3)->(9,4)->(9,5)->(9,6)->(8,6)->(8,7)->(8,8)->(9,8)->(9,9)->(10,9)->(11,9)->(11,8)
27

思路：

 这题是一道经典的广搜，一层一层的搜下去，可以再走的过程中把每个节点的父节点记录下来，这样最后就可以逆着输出。

程序如下：

var
 i,j,k,x,y,px,py,qx,qy,s,last:longint;
 a:array [0..13,0..13] of longint;
 father:array [0..10000] of longint;
 state:array [0..10000,1..3]of longint;
 dx:array [1..4,1..2] of longint;
 procedure print(x:longint);
 begin
  if x=0 then exit;
  inc(s);
  print(father[x]);
  if x<>last then write('(',state[x,1],',',state[x,2],')->')
             else writeln('(',state[x,1],',',state[x,2],')');


 end;
 procedure work;
 var
  head,tail,i,wayn:longint;
 begin
  head:=0;tail:=1;state[1,1]:=px;state[1,2]:=py;
  father[1]:=0;
  repeat
   inc(head);
   for wayn:=1 to 4 do
    begin
    x:=state[head,1]+dx[wayn,1];
    y:=state[head,2]+dx[wayn,2];
    if (x<=i)and(x>=1)and(y<=i)and(y>=1)and(a[x,y]=0) then
                                                       begin
                                                        inc(tail);
                                                        father[tail]:=head;
                                                        state[tail,1]:=state[head,1]+dx[wayn,1];
                                                        state[tail,2]:=state[head,2]+dx[wayn,2];
                                                        a[state[tail,1],state[tail,2]]:=1;
                                                        if (state[tail,1]=qx)and(state[tail,2]=qy) then
                                                           begin

                                                              s:=0;
                                                              last:=tail;
                                                                                 print(tail);
                                                                                 writeln(s);
                                                                                 tail:=0;
                                                            end;
                                                       end;
    end;
  until head>=tail;
 end;
begin
 dx[1,1]:=1;dx[1,2]:=0;
 dx[2,1]:=0;dx[2,2]:=-1;
 dx[3,1]:=-1;dx[3,2]:=0;
 dx[4,1]:=0;dx[4,2]:=1;
 readln(i);
 read(px,py,qx,qy);
 for j:=1 to i do
  for k:=1 to i do
   read(a[j,k]);
  work;
end.