帮我用c++写一段最短的代码实现以下问题。有一个人8*8的棋盘,行号列号均为0-7,一个特殊方格的位置是(5,6),给出采用l形骨牌覆盖其他全部方格的一种方案

要解决这个问题，我们可以使用经典的递归算法来覆盖一个带有特殊方格的8x8棋盘。L形骨牌可以覆盖3个方格，而8x8棋盘共有64个方格，其中有一个特殊方格，所以需要21个L形骨牌来覆盖剩下的63个方格。 以下是一个C++程序，实现了这个算法： ```cpp #include <iostream> #include <vector> const int BOARD_SIZE = 8; const int TILES = 21; // 21 L形骨牌 int board[BOARD_SIZE][BOARD_SIZE] = {0}; // Function to print the board void printBoard() { for (int i = 0; i < BOARD_SIZE; i++) { for (int j = 0; j < BOARD_SIZE; j++) { std::cout << board[i][j] << " "; } std::cout << std::endl; } } // Function to cover the board with L-shaped tiles void cover(int topRow, int leftCol, int bottomRow, int rightCol, int specialRow, int specialCol, int tile) { if (topRow == bottomRow && leftCol == rightCol) { return; } int size = bottomRow - topRow + 1; int halfSize = size / 2; int centerRow = topRow + halfSize - 1; int centerCol = leftCol + halfSize - 1; int nextTile = tile + 1; // Determine the quadrant of the special square if (specialRow <= centerRow && specialCol <= centerCol) { // Special square is in the top-left quadrant board[centerRow][centerCol + 1] = tile; board[centerRow + 1][centerCol] = tile; board[centerRow + 1][centerCol + 1] = tile; cover(topRow, leftCol, centerRow, centerCol, specialRow, specialCol, nextTile); cover(topRow, centerCol + 1, centerRow, rightCol, centerRow, centerCol + 1, nextTile + 1); cover(centerRow + 1, leftCol, bottomRow, centerCol, centerRow + 1, centerCol, nextTile + 2); cover(centerRow + 1, centerCol + 1, bottomRow, rightCol, centerRow + 1, centerCol + 1, nextTile + 3); } else if (specialRow <= centerRow && specialCol > centerCol) { // Special square is in the top-right quadrant board[centerRow][centerCol] = tile; board[centerRow + 1][centerCol] = tile; board[centerRow + 1][centerCol + 1] = tile; cover(topRow, leftCol, centerRow, centerCol, centerRow, centerCol, nextTile); cover(topRow, centerCol + 1, centerRow, rightCol, specialRow, specialCol, nextTile + 1); cover(centerRow + 1, leftCol, bottomRow, centerCol, centerRow + 1, centerCol, nextTile + 2); cover(centerRow + 1, centerCol + 1, bottomRow, rightCol, centerRow + 1, centerCol + 1, nextTile + 3); } else if (specialRow > centerRow && specialCol <= centerCol) { // Special square is in the bottom-left quadrant board[centerRow][centerCol] = tile; board[centerRow][centerCol + 1] = tile; board[centerRow + 1][centerCol + 1] = tile; cover(topRow, leftCol, centerRow, centerCol, centerRow, centerCol, nextTile); cover(topRow, centerCol + 1, centerRow, rightCol, centerRow, centerCol + 1, nextTile + 1); cover(centerRow + 1, leftCol, bottomRow, centerCol, specialRow, specialCol, nextTile + 2); cover(centerRow + 1, centerCol + 1, bottomRow, rightCol, centerRow + 1, centerCol + 1, nextTile + 3); } else { // Special square is in the bottom-right quadrant board[centerRow][centerCol] = tile; board[centerRow][centerCol + 1] = tile; board[centerRow + 1][centerCol] = tile; cover(topRow, leftCol, centerRow, centerCol, centerRow, centerCol, nextTile); cover(topRow, centerCol + 1, centerRow, rightCol, centerRow, centerCol + 1, nextTile + 1); cover(centerRow + 1, leftCol, bottomRow, centerCol, centerRow + 1, centerCol, nextTile + 2); cover(centerRow + 1, centerCol + 1, bottomRow, rightCol, specialRow, specialCol, nextTile + 3); } } int main() { int specialRow = 5, specialCol = 6; cover(0, 0, BOARD_SIZE - 1, BOARD_SIZE - 1, specialRow, specialCol, 1); printBoard(); return 0; } ``` 解释 1. `board`数组：用于存储棋盘的状态，每个位置的值表示覆盖该位置的L形骨牌的编号。 2. `printBoard`函数：用于打印棋盘的状态。 3. `cover`函数：递归地覆盖棋盘。它将棋盘分成四个象限，并根据特殊方格的位置决定如何放置L形骨牌。 4. `main`函数：初始化特殊方格的位置，并调用`cover`函数开始覆盖棋盘，最后打印结果。 运行这个程序，你会看到一个8x8的棋盘，其中每个数字表示覆盖该位置的L形骨牌的编号。特殊方格的位置为(5,6)，其值为0。 [2024-10-16 15:50:50 | AI写代码神器 | 1331点数解答]