# 52.n-queens-ii

## Statement

• Difficulty: Hard
• Tag: `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。

``````输入：n = 4

``````

``````输入：n = 1

``````

• `1 <= n <= 9`

• Difficulty: Hard
• Tag: `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 the number of distinct solutions to the n-queens puzzle.

Example 1:

``````Input: n = 4
Output: 2
Explanation: There are two distinct solutions to the 4-queens puzzle as shown.
``````

Example 2:

``````Input: n = 1
Output: 1
``````

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

class Solution {
public:
static const int N = 1e2 + 10;

int n, res;
int g[N][N];

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

int totalNQueens(int n) {
this->n = n;
res = 0;
memset(g, 0, sizeof g);

dfs(1);

return res;
}
};

#ifdef LOCAL

int main() {
return 0;
}

#endif
``````