回溯法 n皇后 python 解法一

2024年1月29日 421次阅读 来源: 回溯法

输出所有解

global N
N = 4


def printSolution(board):
    for i in range(N):
        for j in range(N):
            print(board[i][j], end="")
        print()
    print()


def isSafe(board, row, col):
    for i in range(col):
        if board[row][i] == 1:
            return False
    for i, j in zip(range(row, -1, -1), range(col, -1, -1)):
        if board[i][j] == 1:
            return False
    for i, j in zip(range(row, N, 1), range(col, -1, -1)):
        if board[i][j] == 1:
            return False
    return True


def solveNQUtil(board, col):
    if col >= N:
        printSolution(board)
    else:
    for i in range(N):
        if isSafe(board, i, col):
            board[i][col] = 1
            solveNQUtil(board, col + 1)
            board[i][col] = 0
    return False


def solveNQ():
    # 写成这样更具适应性
    # board = [[0 for i in range(N)] for j in range(N)]
    board = [[0, 0, 0, 0],
             [0, 0, 0, 0],
             [0, 0, 0, 0],
             [0, 0, 0, 0]
             ]

    if not solveNQUtil(board, 0):
        print("Solution does not exist")
        return False

    printSolution(board)
    return True


if __name__ == "__main__":
    solveNQ()

# output
# 0010
# 1000
# 0001
# 0100

# 0100
# 0001
# 1000
# 0010

若只需要找出一个解，则稍稍改一下即可

global N
N = 4


def printSolution(board):
    for i in range(N):
        for j in range(N):
            print(board[i][j], end="")
        print()
    print()


def isSafe(board, row, col):
    for i in range(col):
        if board[row][i] == 1:
            return False
    # 检查对角线（左上角）是否有冲突
    for i, j in zip(range(row, -1, -1), range(col, -1, -1)):
        if board[i][j] == 1:
            return False
    # 检查对角线（左下角）是否有冲突
    for i, j in zip(range(row, N, 1), range(col, -1, -1)):
        if board[i][j] == 1:
            return False
    return True


def solveNQUtil(board, col):
    if col >= N:
        # printSolution(board)
        return True
    for i in range(N):
        if isSafe(board, i, col):
            board[i][col] = 1
            # 有变动的地方
            if solveNQUtil(board, col + 1):
                return True
            board[i][col] = 0
    return False


def solveNQ():
    # board = [[0 for i in range(N)] for j in range(N)]
    board = [[0, 0, 0, 0],
             [0, 0, 0, 0],
             [0, 0, 0, 0],
             [0, 0, 0, 0]
             ]

    if not solveNQUtil(board, 0):
        print("Solution does not exist")
        return False

    printSolution(board)
    return True


if __name__ == "__main__":
    solveNQ()

# output
# 0010
# 1000
# 0001
# 0100
    原文作者：回溯法
    原文地址: https://blog.csdn.net/normol/article/details/78654822
    本文转自网络文章，转载此文章仅为分享知识，如有侵权，请联系博主进行删除。
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