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			63 lines
		
	
	
		
			1.9 KiB
		
	
	
	
		
			Python
		
	
	
	
	
	
			
		
		
	
	
			63 lines
		
	
	
		
			1.9 KiB
		
	
	
	
		
			Python
		
	
	
	
	
	
"""
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File: n_queens.py
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Created Time: 2023-04-26
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Author: krahets (krahets@163.com)
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"""
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def backtrack(
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    row: int,
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    n: int,
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    state: list[list[str]],
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    res: list[list[list[str]]],
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    cols: list[bool],
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    diags1: list[bool],
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    diags2: list[bool],
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):
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    """回溯算法:n 皇后"""
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    # 当放置完所有行时,记录解
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    if row == n:
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        res.append([list(row) for row in state])
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        return
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    # 遍历所有列
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    for col in range(n):
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        # 计算该格子对应的主对角线和次对角线
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        diag1 = row - col + n - 1
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        diag2 = row + col
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        # 剪枝:不允许该格子所在列、主对角线、次对角线上存在皇后
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        if not cols[col] and not diags1[diag1] and not diags2[diag2]:
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            # 尝试:将皇后放置在该格子
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            state[row][col] = "Q"
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            cols[col] = diags1[diag1] = diags2[diag2] = True
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            # 放置下一行
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            backtrack(row + 1, n, state, res, cols, diags1, diags2)
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            # 回退:将该格子恢复为空位
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            state[row][col] = "#"
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            cols[col] = diags1[diag1] = diags2[diag2] = False
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def n_queens(n: int) -> list[list[list[str]]]:
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    """求解 n 皇后"""
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    # 初始化 n*n 大小的棋盘,其中 'Q' 代表皇后,'#' 代表空位
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    state = [["#" for _ in range(n)] for _ in range(n)]
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    cols = [False] * n  # 记录列是否有皇后
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    diags1 = [False] * (2 * n - 1)  # 记录主对角线上是否有皇后
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    diags2 = [False] * (2 * n - 1)  # 记录次对角线上是否有皇后
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    res = []
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    backtrack(0, n, state, res, cols, diags1, diags2)
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    return res
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"""Driver Code"""
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if __name__ == "__main__":
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    n = 4
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    res = n_queens(n)
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    print(f"输入棋盘长宽为 {n}")
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    print(f"皇后放置方案共有 {len(res)} 种")
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    for state in res:
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        print("--------------------")
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        for row in state:
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            print(row)
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