程序设计方法学.docx

双人对弈算法：通过设置一个全局变量bool color，初始化为true，每一次落子就对color的值进行变换，通过color的值对子的颜色进行修改。代码中还包含了对落子的颜色记录到一个19*19的int型矩阵中以及对以落子的点的保护bool color = true;int piece1919=0;void CWzqView:vDrawPiece(CPoint pointBoard, int nType, CDC* pDC)CRgn rgnPiece;CPoint pointView = ptConvertBP2VP(pointBoard);rgnPiece.CreateEllipticRgn( pointView.x - WIDTH_Piece,pointView.y - WIDTH_Piece,pointView.x + WIDTH_Piece,pointView.y + WIDTH_Piece);CBrush brush;if (color = true)brush.CreateSolidBrush(COLOR_PieceBlack);piecepointBoard.xpointBoard.y = 1;color = false;elsebrush.CreateSolidBrush(COLOR_PieceWhite);piecepointBoard.xpointBoard.y = 2;color = true;pDC-FillRgn(&rgnPiece, &brush);rgnPiece.DeleteObject();/ CWzqView:vDrawPiece判断输赢因为五子棋如果获胜，一定跟刚落的子有关。可根据这一原则对最近刚落下的子进行4个方向的循环扫描，如果有连成5个子就获胜，以横向为例进行了注释，其他三个方向的扫描只是对横向扫描的部分判断条件进行修改。/ draw piecept = pDoc-ptGetPiecePos();if (pt.x = 0) & (pt.y = 0) & piecept.xpt.y = 0)vDrawPiece(pt, 0, pDC);/横向扫描int xx = pt.x, yy = pt.y;for (int i = 4;i 0;i-)if (-xx -1)/ 判断是否在最左边以右if (piecexxyy != piecept.xpt.y)/判断左边的子是否颜色不同xx = pt.x;yy = pt.y;for (int j = i;j 0;j-)/向右扫描，并将i的值赋给jif (+xx 0;j-)if (+xx 0;i-)if (-yy -1)if (piecexxyy != piecept.xpt.y)xx = pt.x;yy = pt.y;for (int j = i;j 0;j-)if (+yy 0;j-)if (+yy 0;i-)if (-xx -1&-yy-1)if (piecexxyy != piecept.xpt.y)xx = pt.x;yy = pt.y;for (int j = i;j 0;j-)if (+xx 19&+yy 0;j-)if (+xx 19&+yy 0;i-)if (-xx -1&+yy 0;j-)if (+xx -1)if (piecexxyy != piecept.xpt.y)break;elsebreak;if (j = 1 & piecexxyy = piecept.xpt.y)MessageBox(胜利, 提示!);break;break;elsexx = pt.x;yy = pt.y;for (int j = i;j 0;j-)if (+xx -1)if (piecexxyy != piecept.xpt.y)break;elsebreak;if (j = 1 & piecexxyy = piecept.xpt.y)MessageBox(胜利, 提示!);