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51.n-queens

Statement

Metadata
  • Link: N 皇后
  • Difficulty: Hard
  • Tag: 数组 回溯

n 皇后问题 研究的是如何将 n 个皇后放置在 n×n 的棋盘上，并且使皇后彼此之间不能相互攻击。

给你一个整数 n ，返回所有不同的 n 皇后问题 的解决方案。

每一种解法包含一个不同的 n 皇后问题 的棋子放置方案，该方案中 'Q' 和 '.' 分别代表了皇后和空位。

 

示例 1：

输入：n = 4
输出：[[".Q..","…Q","Q…","..Q."],["..Q.","Q…","…Q",".Q.."]]
解释：如上图所示，4 皇后问题存在两个不同的解法。

示例 2：

输入：n = 1
输出：[["Q"]]

 

提示：

  • 1 <= n <= 9

Metadata
  • Link: N-Queens
  • Difficulty: Hard
  • Tag: Array Backtracking

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.

Given an integer n, return all distinct solutions to the n-queens puzzle. You may return the answer in any order.

Each solution contains a distinct board configuration of the n-queens' placement, where 'Q' and '.' both indicate a queen and an empty space, respectively.

 

Example 1:

Input: n = 4
Output: [[".Q..","…Q","Q…","..Q."],["..Q.","Q…","…Q",".Q.."]]
Explanation: There exist two distinct solutions to the 4-queens puzzle as shown above

Example 2:

Input: n = 1
Output: [["Q"]]

 

Constraints:

  • 1 <= n <= 9

Solution

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

#define endl "\n"
#define fi first
#define se second
#define all(x) begin(x), end(x)
#define rall rbegin(a), rend(a)
#define bitcnt(x) (__builtin_popcountll(x))
#define complete_unique(a) a.erase(unique(begin(a), end(a)), end(a))
#define mst(x, a) memset(x, a, sizeof(x))
#define MP make_pair

using ll = long long;
using ull = unsigned long long;
using db = double;
using ld = long double;
using VLL = std::vector<ll>;
using VI = std::vector<int>;
using PII = std::pair<int, int>;
using PLL = std::pair<ll, ll>;

using namespace __gnu_pbds;
using namespace std;
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
const ll mod = 1e9 + 7;

template <typename T, typename S>
inline bool chmax(T &a, const S &b) {
    return a < b ? a = b, 1 : 0;
}

template <typename T, typename S>
inline bool chmin(T &a, const S &b) {
    return a > b ? a = b, 1 : 0;
}

#ifdef LOCAL
#include <debug.hpp>
#else
#define dbg(...)
#endif
// head

class Solution {
public:
    int n;
    vector<vector<string>> res;

    vector<string> toString(const vector<vector<int>> &g) {
        auto cur_res = vector<string>();
        for (int i = 1; i <= n; i++) {
            string s = "";
            for (int j = 1; j <= n; j++) {
                if (g[i][j]) {
                    s += "Q";
                } else {
                    s += ".";
                }
            }
            cur_res.push_back(s);
        }

        return cur_res;
    }

    bool ok(int x, int y, const vector<vector<int>> &g) {
        for (int i = 1; i <= n; ++i) {
            for (int j = 1; j <= n; ++j) {
                if (!(x == i && y == j)) {
                    if (g[i][j] == 1) {
                        if (i == x)
                            return false;
                        if (j == y)
                            return false;
                        if (i + j == x + y)
                            return false;
                        if (i - j == x - y)
                            return false;
                    }
                }
            }
        }
        return true;
    }

    void dfs(int cur, vector<vector<int>> &g) {
        if (cur == n + 1) {
            res.push_back(toString(g));
            return;
        }

        for (int i = 1; i <= n; ++i) {
            if (ok(cur, i, g)) {
                g[cur][i] = 1;
                dfs(cur + 1, g);
                g[cur][i] = 0;
            }
        }
    }

    vector<vector<string>> solveNQueens(int n) {
        this->n = n;
        auto g = vector<vector<int>>(n + 1, vector<int>(n + 1, 0));
        dfs(1, g);
        return res;
    }
};

#ifdef LOCAL

int main() {
    return 0;
}

#endif
最后更新: October 11, 2023
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