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* Revised the book * Update the book with the second revised edition * Revise base on the manuscript of the first edition
62 lines
1.9 KiB
Python
62 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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