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找一找教程网。
