经过一个月的调试后我编写了这个程序，我终于开始工作了，但它只打印1问题的8解决方案，是否有人知道我能做些什么来制作它打印所有解决方案？代码将会有所帮助，但如果您只是指出要更改的内容或添加的内容，我也可以使用它。
import java.util.Scanner;
public class Queens
{
// squares per row or column
public static final int BOARD_SIZE = 8;
// used to indicate an empty square
public static final int EMPTY = 0;
// used to indicate square contains a queen
public static final int QUEEN = 1;
private int board[][]; // chess board
public Queens() {
// -------------------------------------------------
// Constructor: Creates an empty square board.
// -------------------------------------------------
board = new int[BOARD_SIZE][BOARD_SIZE];
} // end constructor
public void clearBoard() {
// -------------------------------------------------
// Clears the board.
// Precondition: None.
// Postcondition: Sets all squares to EMPTY.
// -------------------------------------------------
for(int j = 1; j < 8; j++)
{
for(int k = 1; k < 8; k++) //Sets every column in this row to 0
{
board[j][k] = 0;
}
//moves on to next row and repeats
}
} // end clearBoard
public void displayBoard() {
// -------------------------------------------------
// Displays the board.
// Precondition: None.
// Postcondition: Board is written to standard
// output; zero is an EMPTY square, one is a square
// containing a queen (QUEEN).
// -------------------------------------------------
placeQueens(1);
int N = board.length;
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
if (board[i][j] == 1)
{
System.out.print("Q ");
}
else
{
System.out.print("_|");
}
}
System.out.println();
}
} // end displayBoard
public boolean placeQueens(int column) {
// -------------------------------------------------
// Places queens in columns of the board beginning
// at the column specified.
// Precondition: Queens are placed correctly in
// columns 1 through column-1.
// Postcondition: If a solution is found, each
// column of the board contains one queen and method
// returns true; otherwise, returns false (no
// solution exists for a queen anywhere in column
// specified).
// -------------------------------------------------
if (column > BOARD_SIZE) {
return true; // base case
}
else {
boolean queenPlaced = false;
int row = 1; // number of square in column
while ( !queenPlaced && (row <= BOARD_SIZE) ) {
// if square can be attacked
if (isUnderAttack(row, column)) {
++row; // consider next square in column
} // end if
else { // place queen and consider next column
setQueen(row, column);
queenPlaced = placeQueens(column+1);
// if no queen is possible in next column,
if (!queenPlaced) {
// backtrack: remove queen placed earlier
// and try next square in column
removeQueen(row, column);
++row;
} // end if
} // end if
} // end while
return queenPlaced;
} // end if
} // end placeQueens
private void setQueen(int row, int column) {
// --------------------------------------------------
// Sets a queen at square indicated by row and
// column.
// Precondition: None.
// Postcondition: Sets the square on the board in a
// given row and column to QUEEN.
// --------------------------------------------------
row = index(row);
column = index(column);
board[row][column] = 1; //Queen placed on square
} // end setQueen
private void removeQueen(int row, int column) {
// --------------------------------------------------
// Removes a queen at square indicated by row and
// column.
// Precondition: None.
// Postcondition: Sets the square on the board in a
// given row and column to EMPTY.
// --------------------------------------------------
column = index(column);
for(int x = 0; x < 8 ; x++)
{
if(board[x][column] == 1)
{
board[x][column] = 0;
x = 8;
}
}
} // end removeQueen
private boolean isUnderAttack(int row, int column) {
// --------------------------------------------------
// Determines whether the square on the board at a
// given row and column is under attack by any queens
// in the columns 1 through column-1.
// Precondition: Each column between 1 and column-1
// has a queen placed in a square at a specific row.
// None of these queens can be attacked by any other
// queen.
// Postcondition: If the designated square is under
// attack, returns true; otherwise, returns false.
// --------------------------------------------------
//Taking 1-8 & returning 0-7 to suite array
row = index(row);
column = index(column);
//Checks the rows & columns
//Rows
for(int i = 0; i < column && i < 8 && row < 8; i++)
{
//If there's a queen in that row, the queen is under attack
if(board[row][i] == 1)
{
return true;
}
}
//Column
for(int j = 0; j < row && j < 8 && column < 8; j++)
{
//If there's a queen in that column, the queen is under attack
if(board[j][column] == 1)
{
return true;
}
}
//Check diagonals
for(int i = row, j = column; i >= 0 && j >= 0 && i < 8 && j < 8; i--, j--)
{
//checks upper diagonal
if(board[i][j] == 1)
{
return true;
}
}
for(int i = row, j = column; i < board.length && j >= 0 && i < 8 && j < 8; i++, j--)
{
//checks lower diagonal
if(board[i][j] == 1)
{
return true;
}
}
//At this point the Queen is not being attacked
return false;
} // end isUnderAttack
private int index(int number) {
// --------------------------------------------------
// Returns the array index that corresponds to
// a row or column number.
// Precondition: 1 <= number <= BOARD_SIZE.
// Postcondition: Returns adjusted index value.
// --------------------------------------------------
return number - 1;
}// end index
public static void main(String[] args)
{
Queens eight = new Queens();
eight.displayBoard();
}
} // end Queens
答案 0 :(得分：1)
displayBoard 是您的驾驶习惯;在显示一个解决方案后，不要让它停止，只要 placeQueens 找到 new 解决方案，就将其包装在一个循环中。
这意味着您需要调整 placeQueens 以继续上一个电路板状态。它已经在很大程度上做到了这一点;你只需要处理它击中最后一列的情况。例如，将第8个女王移动到一个方格并继续从你离开的地方继续 - 或者回到第7个女王的下一个合法位置（因为你知道第8个没有其他合法地点）。
当你这样做时，你需要稍微改变这两个例程之间的接口，这样placeQueens不仅可以返回每个解决方案，还可以返回全部完成条件。这告诉 displayBoard 打破循环（你添加的新包装器）。
这足以让你感动吗？
编辑后＆＃34;不是真的＆＃34;评论......
也许最容易编写的包装器将在 displayBoard 中。在顶部，您有 placeQueens（1），而不是使用
col = 1
while(placeQueens(col)) {
... print the board as usual
... remove 8th queen; mark its position as unusable (say, with a value of -1)
col = 8
}
调整placeQueens，使其从停止的地方开始拾取：它会想要将第8个女王放在同一个地方。当发现该点被标记为不可用时，重置该标记并回溯到第7个女王。此操作将继续并找到所有解决方案。
有更简洁的方法可以做到这一点，但我认为这个方法足以保留您现有的组织。理想情况下，您有一个循环执行布局和打印过程的驱动程序，但这主要是命名和上层组织的问题。
这个能让你移动得足够好吗？