【LeetCode】N-Queens II

最新推荐文章于 2022-06-06 10:35:33 发布

ice4026

最新推荐文章于 2022-06-06 10:35:33 发布

阅读量501

点赞数

CC 4.0 BY-SA版权

分类专栏：算法学习【LeetCode】

本文链接：https://blog.youkuaiyun.com/ice4026/article/details/31232037

算法学习同时被 2 个专栏收录

104 篇文章

订阅专栏

【LeetCode】

100 篇文章

订阅专栏

本文介绍了一种解决N皇后问题的方法，不仅提供了一种递归解决方案来计算所有可能的棋盘配置数量，还展示了一种使用位操作进行快速计算的高级技巧。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目描述：

Follow up for N-Queens problem.

Now, instead outputting board configurations, return the total number of distinct solutions.

本来我觉得会有dp解法，想了很久没想出来，就拿 N-Queens改了下，然后就AC了……都是泪……

class Solution {
public:
	int count;
	int N;
	int totalNQueens(int n) {
		N = n;
		count = 0;
		vector<int> rec;
		solve(n, rec);
		return count;
	}
	void solve(int n, vector<int> &rec){
		if (!n){
			count++;
			return;
		}
		int i = rec.size();
		for (int j = 0; j < N; j++){
			int m(0);
			for (m = 0; m < rec.size(); m++)
			if (rec[m] == j || m + rec[m] == i + j || m - rec[m] == i - j)
				break;
			if (m < rec.size())
				continue;
			rec.push_back(j);
			solve(n - 1, rec);
			rec.pop_back();
		}
	}
};

还有个在讨论区看到的很神奇的解法记录下，后面再研究……

I found this answer on the Internet. And I spent a while to figure how it works, I put the main idea on the comment. Here it is!

And a picture to illustrate how the bits projection works. enter image description here

class Solution {
    public:
        /* backtrace program using bit-wise operation to speed up calculation.
         * 'limit' is all '1's.
         * 'h'  is the bits all the queens vertically projected on a row. If h==limit, then it's done, answer++.
         * 'r'   is the bits all the queens anti-diagonally projected on a row.
         * 'l'   is the bits all the queens diagonally projected on a row.
         * h|r|l  is all the occupied bits. Then pos = limit & (~(h|r|l)) is all the free positions.
         * p = pos & (-pos)  gives the right most '1'. pos -= p means we will place a queen on this bit 
         *                             represented by p.
         * 'h+p'  means one more queue vertically projected on next row.
         * '(r+p)<<1'  means one more queue anti-diagonally projected on next row. Because we are
         *                   moving to next row and the projection is skew from right to left, we have to 
         *                   shift left one position after moved to next row.
         * '(l+p)>>1'  means one more queue diagonally projected on next row. Because we are 
         *                  moving to next row and the projection is skew from left to right, we have to 
         *                  shift right one position after moved to next row.
         */
        int totalNQueens(int n) {
            ans = 0;
            limit = (1<<n) - 1;
            dfs(0, 0, 0);
            return ans;
        }
        void dfs(int h, int r, int l) {
            if (h == limit) {
                ans++;
                return;
            }
            int pos = limit & (~(h|r|l));
            while (pos) {
                int p = pos & (-pos);
                pos -= p;
                dfs(h+p, (r+p)<<1, (l+p)>>1);
            }
        }
        int ans, limit;
};