判断五子棋获胜方的一个算法

来源：互联网发布：花刺软件编辑：程序博客网时间：2024/04/29 00:29

int CMainWindow::IsWinner()
{
struct LINE
{
  int topleft,topright,left,top;
};
LINE XGridLine[10][10];
LINE OGridLine[10][10];
for(int i=0;i<10;i++)
  for(int j=0;j<10;j++)
  {
   XGridLine[i][j].topleft=
    XGridLine[i][j].topright=
    XGridLine[i][j].left=
    XGridLine[i][j].top=0;
   OGridLine[i][j].topleft=
    OGridLine[i][j].topright=
    OGridLine[i][j].left=
    OGridLine[i][j].top=0;
  }
for(i=0;i<10;i++)
  for(int j=0;j<10;j++)
  {
   if(m_nGameGrid[i][j]==EX)
   {
    if(j-1>=0)
     XGridLine[i][j].left=XGridLine[i][j-1].left+1;
    else
     XGridLine[i][j].left=1;
    if(i-1>=0)
     XGridLine[i][j].top=XGridLine[i-1][j].top+1;
    else
     XGridLine[i][j].top=1;
    if(i-1>=0&&j-1>=0)
     XGridLine[i][j].topleft=XGridLine[i-1][j-1].topleft+1;
    else
     XGridLine[i][j].topleft=1;
    if(i-1>=0&&j+1<10)
     XGridLine[i][j].topright=XGridLine[i-1][j+1].topright+1;
    else
     XGridLine[i][j].topright=1;
   }
   if(m_nGameGrid[i][j]==OH)
   {
    if(j-1>=0)
     OGridLine[i][j].left=OGridLine[i][j-1].left+1;
    else
     OGridLine[i][j].left=1;
    if(i-1>=0)
     OGridLine[i][j].top=OGridLine[i-1][j].top+1;
    else
     OGridLine[i][j].top=1;
    if(i-1>=0&&j-1>=0)
     OGridLine[i][j].topleft=OGridLine[i-1][j-1].topleft+1;
    else
     OGridLine[i][j].topleft=1;
    if(i-1>=0&&j+1<10)
     OGridLine[i][j].topright=OGridLine[i-1][j+1].topright+1;
    else
     OGridLine[i][j].topright=1;
   }
   if(XGridLine[i][j].left>=5||XGridLine[i][j].top>=5
    ||XGridLine[i][j].topleft>=5||XGridLine[i][j].topright>=5)
   { return EX; }
   if(OGridLine[i][j].left>=5||OGridLine[i][j].top>=5
    ||OGridLine[i][j].topleft>=5||OGridLine[i][j].topright>=5)
   { return OH; }
  }
  return 0;
}