本文实例为大家分享了MFC实现连连看游戏消子算法的具体代码，供大家参考，具体内容如下

两个位置的图片能否消除，有三种情况：

1.一条直线连接，这种也是最简单的一种消除方法

bool LinkInLine(CPoint p1, CPoint p2) { conner1.x = conner1.y = -1; // 记录拐点位置 conner2.x = conner2.y = -1; BOOL b = true; if (p1.y == p2.y) // 两个点再同一行 { int min_x = min(p1.x, p2.x); int max_x = max(p1.x, p2.x); for (int i = min_x+1; i < max_x; i++) { if (game->map[i][p1.y] != 0) { b = false; } } } else if (p1.x == p2.x) // 在同一列 { int min_y = min(p1.y, p2.y); int max_y = max(p1.y, p2.y); for (int i = min_y + 1; i < max_y; i++) { if (game->map[p1.x][i] != 0) { b = false; } } } else // 不在同一直线 { b = false; } return b;}

2.两条直线消除，即经过一个拐点。

两个顶点经过两条直线连接有两种情况，即两个拐点分两种情况。

bool OneCornerLink(CPoint p1, CPoint p2) { conner1.x = conner1.y = -1; conner2.x = conner2.y = -1; int min_x = min(p1.x, p2.x); int max_x = max(p1.x, p2.x); int min_y = min(p1.y, p2.y); int max_y = max(p1.y, p2.y); // 拐点1 int x1 = p1.x; int y1 = p2.y; //拐点2 int x2 = p2.x; int y2 = p1.y; BOOL b = true; if (game->map[x1][y1] != 0 && game->map[x2][y2] != 0) { b = false; } else { if (game->map[x1][y1] == 0) // 拐点1位置无图片 { for (int i = min_x + 1; i < max_x; i++) { if (game->map[i][y1] != 0) { b = false; break; } } for (int i = min_y + 1; i < max_y; i++) { if (game->map[x1][i] != 0) { b = false; break; } } if (b) { conner1.x = x1; conner1.y = y1; return b; } } if (game->map[x2][y2] == 0) // 拐点2位置无图片 { b = true; for (int i = min_x + 1; i < max_x; i++) { if (game->map[i][y2] != 0) { b = false; break; } } for (int i = min_y + 1; i < max_y; i++) { if (game->map[x2][i] != 0) { b = false; break; } } if (b) { conner1.x = x2; conner1.y = y2; return b; } } } return b;}

3.三条直线消除，即经过两个拐点。

这是可以通过横向扫描和纵向扫描，扫描的时候可以得到连个拐点，判断两个顶点经过这两个拐点后是否能消除

bool TwoCornerLink(CPoint p1, CPoint p2) { conner1.x = conner1.y = -1; conner2.x = conner2.y = -1; int min_x = min(p1.x, p2.x); int max_x = max(p1.x, p2.x); int min_y = min(p1.y, p2.y); int max_y = max(p1.y, p2.y); bool b; for (int i = 0; i < MAX_Y; i++) // 扫描行 { b = true; if (game->map[p1.x][i] == 0 && game->map[p2.x][i] == 0) // 两个拐点位置无图片 { for (int j = min_x + 1; j < max_x; j++) // 判断连个拐点之间是否可以连接 { if (game->map[j][i] != 0) { b = false; break; } } if (b) { int temp_max = max(p1.y, i); int temp_min = min(p1.y, i); for (int j = temp_min + 1; j < temp_max; j++) // 判断p1和它所对应的拐点之间是否可以连接 { if (game->map[p1.x][j] != 0) { b = false; break; } } } if (b) { int temp_max = max(p2.y, i); int temp_min = min(p2.y, i); for (int j = temp_min + 1; j < temp_max; j++) // 判断p2和它所对应的拐点之间是否可以连接 { for (int j = temp_min + 1; j < temp_max; j++) { if (game->map[p2.x][j] != 0) { b = false; break; } } } } if (b) // 如果存在路线，返回true { conner1.x = p1.x; conner1.y = i; conner2.x = p2.x; conner2.y = i; return b; } } }// 扫描行结束 for (int i = 0; i < MAX_X; i++) // 扫描列 { b = true; if (game->map[i][p1.y] == 0 && game->map[i][p2.y] == 0) // 连个拐点位置无图片 { for (int j = min_y + 1; j < max_y; j++) // 两个拐点之间是否可以连接 { if (game->map[i][j] != 0) { b = false; break; } } if (b) { int temp_max = max(i, p1.x); int temp_min = min(i, p1.x); for (int j = temp_min + 1; j < temp_max; j++) // 判断p1和它所对应的拐点之间是否可以连接 { if (game->map[j][p1.y] != 0) { b = false; break; } } } if (b) { int temp_max = max(p2.x, i); int temp_min = min(p2.x, i); for (int j = temp_min + 1; j < temp_max; j++) { if (game->map[j][p2.y] != 0) { b = false; break; } } } if (b) // 如果存在路线，返回true { conner1.y = p1.y; conner1.x = i; conner2.y = p2.y; conner2.x = i; return b; } } } // 扫描列结束 return b;}

完整源码已上传至我的GitHub

以上就是本文的全部内容，希望对大家的学习有所帮助，也希望大家多多支持。