2n皇后问题

问题：

思路：
点击查看【bilibili】

N皇后代码

def dfs(chessboard, nums, row, n):
    if row == n:
        ans.append(nums[:])
        return
    for col in range(n): # 对于某行的各个列
        if chessboard[row][col] != 0:
            nums[row] = col # 第row行皇后所在的列数
            if valid(nums, row):
                dfs(chessboard, nums, row+1, n)
def valid(nums, row):
    for i in range(row): # 第row行前面的行
        # 不在同一列
        # 不在对角线上（2， 2）...
        if nums[i] == nums[row] or abs(nums[i]-nums[row]) == abs(row-i):
            return False
    return True
n  = int(input())
chessboard = [[int(i) for i in input().split()] for _ in range(n)]
# print(chessboard)
ans = []
nums = [None for _ in range(n)]
dfs(chessboard, nums, 0, n) # 当前第0行，一共n行
print(len(ans))
print(ans)

2N皇后，简而言之就是有两个皇后，黑皇后和白皇后。

def dfs(chessboard, nums, row, n):
    if row == n:
        ans.append(nums[:])
        return
    for col_white in range(n): # 对于某行的各个列
        if chessboard[row][col_white] != 0:
            nums[0][row] = col_white # 第row行皇后所在的列数
            for col_black in range(n):
                if chessboard[row][col_black] == 0 or col_white == col_black:
                    continue
                nums[1][row] = col_black
                if valid(nums, row):
                    dfs(chessboard, nums, row+1, n)
def valid(nums, row):
    for color in range(2):
        tmp = nums[color]
        for i in range(row): # 第row行前面的行
            # 不在同一列
            # 不在对角线上（2， 2）...
            if tmp[i] == tmp[row] or abs(tmp[i]-tmp[row]) == abs(row-i):
                return False
    return True
n  = int(input())
chessboard = [[int(i) for i in input().split()] for _ in range(n)]
# print(chessboard)
ans = []
nums = [[None]*n for _ in range(2)] # nums[0]代表白皇后列索引；num[1]代表黑皇后列索引
dfs(chessboard, nums, 0, n) # 当前第0行，一共n行
print(len(ans))
print(ans)

ans打印有问题！！