Content

有一個 \(n\times m\) 的拼圖，擺上了幾塊俄羅斯方塊圖形。已知這些圖形可能包含以下這五種（可以旋轉），求出下列型別的俄羅斯方塊圖形數量。

資料範圍：\(1\leqslant n,m\leqslant 10\)。

Solution

像我這樣菜的人，這種題目只有一種方法：暴力判斷。

各位玩過俄羅斯方塊的都知道，上面 \(5\) 種圖形中，第一種圖形無論怎麼旋轉都是一樣的，第二、三、四種都可以通過旋轉得到兩種不同的圖形，第五種可以通過旋轉得到 \(4\) 種不同的圖形。具體是什麼樣的想必各位都能夠想象得出來。

於是，我們只需要對這總共 \(1+2\times 3+4=11\)

然而最煩人的就是打程式碼的過程……

Code

僅搬出判斷五種不同型別的圖形的函式 \(\texttt{judge1}\sim\texttt{judge5}\)，畢竟這才是整個程式碼的精髓，也是我打得最要命的地方……

inline bool judge1(int i, int j) {return a[i][j] != '.' && (a[i][j] == a[i + 1][j] && a[i][j] == a[i + 1][j + 1] && a[i][j] == a[i][j + 1]);}
inline bool judge2(int i, int j) {return a[i][j] != '.' && ((a[i][j] == a[i][j + 1] && a[i][j] == a[i][j + 2] && a[i][j] == a[i][j + 3]) || (a[i][j] == a[i + 1][j] && a[i][j] == a[i + 2][j] && a[i][j] == a[i + 3][j]));}
inline bool judge3(int i, int j) {return a[i][j] != '.' && ((a[i][j] == a[i][j - 1] && a[i][j] == a[i + 1][j - 1] && a[i][j] == a[i + 1][j - 2]) || (a[i][j] == a[i + 1][j] && a[i][j] == a[i + 1][j + 1] && a[i][j] == a[i + 2][j + 1]));}
inline bool judge4(int i, int j) {return a[i][j] != '.' && ((a[i][j] == a[i][j + 1] && a[i][j] == a[i + 1][j + 1] && a[i][j] == a[i + 1][j + 2]) || (a[i][j] == a[i + 1][j] && a[i][j] == a[i + 1][j - 1] && a[i][j] == a[i + 2][j - 1]));}
inline bool judge5(int i, int j) {return a[i][j] != '.' && ((a[i][j] == a[i + 1][j - 1] && a[i][j] == a[i + 1][j] && a[i][j] == a[i + 1][j + 1]) || (a[i][j] == a[i - 1][j - 1] && a[i][j] == a[i][j - 1] && a[i][j] == a[i + 1][j - 1]) || (a[i][j] == a[i - 1][j - 1] && a[i][j] == a[i - 1][j] && a[i][j] == a[i - 1][j + 1]) || (a[i][j] == a[i - 1][j + 1] && a[i][j] == a[i][j + 1] && a[i][j] == a[i + 1][j + 1]));}