30. 老鼠走迷宮

// BCB 程式碼範例

__fastcall TForm1::TForm1(TComponent* Owner)

: TForm(Owner)

{ Memo1->Clear();

move[0].dx = -1;   move[0].dy = 0;

move[1].dx = -1;  move[1].dy = 1;

move[2].dx = 0;    move[2].dy = 1;

move[3].dx = 1;   move[3].dy = 1;

move[4].dx = 1;    move[4].dy = 0;

move[5].dx = 1;   move[5].dy = -1;

move[6].dx = 0;    move[6].dy = -1;

move[7].dx = -1;  move[7].dy = -1;

}

//以上 請寫在 Form1  初始化的地方 (如上)

int u, v, i, j;

position step;

int mark[200][200];

boolean found = false;

Memo1->Lines->Add("m="+IntToStr(m)+" p="+IntToStr(p));

// Memo1->Lines->Add(found);

for (i=1; i<=m; i++)

{   for (j=1; j<=p; j++)

    {   mark[i][j] = maze[i][j];

    }

}

for (i=0; i<=m; i++) { mark[i][0] = 1; mark[i][p+1] = 1; }

for (j=0; i<=p; i++) { mark[0][j] = 1; mark[m+1][j] = 1; }

step.x = 1;

step.y = 1;

step.dir = 2;   // (i,j,dir)=(1,1, 2)

push(step);

while (top != -1)

{    step = pop();

     while ((step.dir <= 7) && (!found))

     {    u = step.x+move[step.dir].dx;

          v = step.y+move[step.dir].dy;

          int x = step.dir;

          Memo1->Lines->Add(".. (u,v,dir)="+

          IntToStr(u)+","+IntToStr(v)+","+IntToStr(x));

          //str = ;

          Memo1->Lines->Add("...come:"+IntToStr(u)+","+

          IntToStr(v)+","+IntToStr(x));

          // (u,v) = 自(i,j)欲嘗試的下一步座標

          if ((u == m) && (v == p))

          {      found = true; //成功找到出口，輸出路徑，可以停止

                  str = "("+IntToStr(m)+","+IntToStr(p)+")";

                 StringGrid1->Cells[p][m] = "&";

                 str = "("+IntToStr(step.x)+","+IntToStr(step.y)+","+IntToStr(x)+")-->"+str;

                 StringGrid1->Cells[step.y][step.x] = "->";

                 while (top != -1)

                 {    step = pop();

                       int x = step.dir;

                       str = "("+IntToStr(step.x)+","+IntToStr(step.y)+","+IntToStr(x)+")-->"+str;

                       StringGrid1->Cells[step.y][step.x] = "->";

                 }

                 Memo1->Lines->Add(str);

          }

          else if ((maze[u][v]==0) && (mark[u][v]==0))

           //(u,v)可以走，且不曾走過

           {      mark[u][v] = 1;

                   StringGrid1->Cells[v][u] = "*";

                   step.dir = step.dir+1;  //?

                   push(step);

                    step.x = u; step.y = v; step.dir = 0;

             }

             else step.dir += 1;

       }

}