N-Queens II 使用回溯很慢 [英] N-Queens II using backtracking is slow

问题描述

n-queens 难题是将 n 个皇后放在 n x n 棋盘上的问题，这样就不会有两个皇后互相攻击.

The n-queens puzzle is the problem of placing n queens on an n x n chessboard such that no two queens attack each other.

给定一个整数 n，返回 n-queens 难题的不同解的数量.

Given an integer n, return the number of distinct solutions to the n-queens puzzle.

https://leetcode.com/problems/n-queens-ii/

我的解决方案:

class Solution:
    def totalNQueens(self, n: int) -> int:
        def genRestricted(restricted, r, c):
            restricted = set(restricted)
            for row in range(n): restricted.add((row, c))
            for col in range(n): restricted.add((r, col))
            movements = [[-1, -1], [-1, 1], [1, -1], [1, 1]]
            for movement in movements:
                row, col = r, c
                while 0 <= row < n and 0 <= col < n:
                    restricted.add((row, col))
                    row += movement[0]
                    col += movement[1]
            return restricted

        def gen(row, col, curCount, restricted):
            count, total_count = curCount, 0

            for r in range(row, n):
                for c in range(col, n):
                    if (r, c) not in restricted:
                        count += 1
                        if count == n: total_count += 1
                        total_count += gen(row + 1, 0, count, genRestricted(restricted, r, c))
                        count -= 1

            return total_count

        return gen(0, 0, 0, set())

它在 n=8 时失败.我不知道为什么，以及如何减少迭代.看来我已经在做尽可能少的迭代了.

It fails at n=8. I can't figure out why, and how to have less iterations. It seems I am already doing the minimum iterations possible.

推荐答案

restricted 集似乎在时间和空间方面都是浪费的.在成功的递归结束时，n 层深度增长到 n^2 大小，这将总复杂度驱动到 O(n^3).它并不是真正需要的.通过查看已经放置的皇后更容易检查正方形的可用性(请原谅国际象棋术语；file 代表垂直，rank 代表水平):

The restricted set seems wasteful, both time- and space-wise. At the end of the successful recursion, n levels deep it grows to n^2 size, which drives the total complexity to O(n^3). And it is not really needed. It is much easier to check the availability of the square by looking at the queens already placed (please forgive the chess lingo; file stand for vertical, and rank for horizontal):

def square_is_safe(file, rank, queens_placed):
    for queen_rank, queen_file in enumerate(queens_placed):
        if queen_file == file:                      # vertical attack
            return false
        if queen_file - file == queen_rank - rank:  # diagonal attack
            return false
        if queen_file - file == rank - queen_rank:  # anti-diagonal attack
            return false
    return true

用于

def place_queen_at_rank(queens_placed, rank):
    if rank == n:
        total_count += 1
        return

    for file in range(0, n):
        if square_is_safe(file, rank, queens_placed):
            queens_placed.append(file)
            place_queen_at_rank(queens_placed, rank + 1)

    queens_placed.pop()

而且还有很大的优化空间.例如，您可能希望对第一级进行特殊处理:由于对称性，您只需要检查它的一半(将执行时间减少 2 倍).

And there is a plenty of room for the optimization. For example, you may want to special-case the first rank: due to a symmetry, you only need to inspect a half of it (cutting execution time by the factor of 2).

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