MFC實現連連看三:消子演算法
阿新 • • 發佈:2019-01-11
兩個位置的圖片能否消除,有三種情況:
1.一條直線連線,這種也是最簡單的一種消除方法
bool LinkInLine(CPoint p1, CPoint p2)
{
conner1.x = conner1.y = -1; // 記錄拐點位置
conner2.x = conner2.y = -1;
BOOL b = true;
if (p1.y == p2.y) // 兩個點再同一行
{
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 // 不在同一直線
{
b = false;
}
return b;
}
2.兩條直線消除,即經過一個拐點。
兩個頂點經過兩條直線連線有兩種情況,即兩個拐點分兩種情況。
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);
// 拐點1
int x1 = p1.x;
int y1 = p2.y;
//拐點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) // 拐點1位置無圖片
{
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) // 拐點2位置無圖片
{
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;
}
3.三條直線消除,即經過兩個拐點。
這是可以通過橫向掃描和縱向掃描,掃描的時候可以得到連個拐點,判斷兩個頂點經過這兩個拐點後是否能消除
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++) // 掃描行
{
b = true;
if (game->map[p1.x][i] == 0 && game->map[p2.x][i] == 0) // 兩個拐點位置無圖片
{
for (int j = min_x + 1; j < max_x; j++) // 判斷連個拐點之間是否可以連線
{
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++) // 判斷p1和它所對應的拐點之間是否可以連線
{
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++) // 判斷p2和它所對應的拐點之間是否可以連線
{
for (int j = temp_min + 1; j < temp_max; j++)
{
if (game->map[p2.x][j] != 0)
{
b = false;
break;
}
}
}
}
if (b) // 如果存在路線,返回true
{
conner1.x = p1.x;
conner1.y = i;
conner2.x = p2.x;
conner2.y = i;
return b;
}
}
}// 掃描行結束
for (int i = 0; i < MAX_X; i++) // 掃描列
{
b = true;
if (game->map[i][p1.y] == 0 && game->map[i][p2.y] == 0) // 連個拐點位置無圖片
{
for (int j = min_y + 1; j < max_y; j++) // 兩個拐點之間是否可以連線
{
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++) // 判斷p1和它所對應的拐點之間是否可以連線
{
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) // 如果存在路線,返回true
{
conner1.y = p1.y;
conner1.x = i;
conner2.y = p2.y;
conner2.x = i;
return b;
}
}
} // 掃描列結束
return b;
}