|
|
/*
|
|
|
* File: n_queens.rs
|
|
|
* Created Time: 2023-07-15
|
|
|
* Author: codingonion (coderonion@gmail.com)
|
|
|
*/
|
|
|
|
|
|
/* 回溯算法：n 皇后 */
|
|
|
fn backtrack(
|
|
|
row: usize,
|
|
|
n: usize,
|
|
|
state: &mut Vec<Vec<String>>,
|
|
|
res: &mut Vec<Vec<Vec<String>>>,
|
|
|
cols: &mut [bool],
|
|
|
diags1: &mut [bool],
|
|
|
diags2: &mut [bool],
|
|
|
) {
|
|
|
// 当放置完所有行时，记录解
|
|
|
if row == n {
|
|
|
res.push(state.clone());
|
|
|
return;
|
|
|
}
|
|
|
// 遍历所有列
|
|
|
for col in 0..n {
|
|
|
// 计算该格子对应的主对角线和次对角线
|
|
|
let diag1 = row + n - 1 - col;
|
|
|
let diag2 = row + col;
|
|
|
// 剪枝：不允许该格子所在列、主对角线、次对角线上存在皇后
|
|
|
if !cols[col] && !diags1[diag1] && !diags2[diag2] {
|
|
|
// 尝试：将皇后放置在该格子
|
|
|
state[row][col] = "Q".into();
|
|
|
(cols[col], diags1[diag1], diags2[diag2]) = (true, true, true);
|
|
|
// 放置下一行
|
|
|
backtrack(row + 1, n, state, res, cols, diags1, diags2);
|
|
|
// 回退：将该格子恢复为空位
|
|
|
state[row][col] = "#".into();
|
|
|
(cols[col], diags1[diag1], diags2[diag2]) = (false, false, false);
|
|
|
}
|
|
|
}
|
|
|
}
|
|
|
|
|
|
/* 求解 n 皇后 */
|
|
|
fn n_queens(n: usize) -> Vec<Vec<Vec<String>>> {
|
|
|
// 初始化 n*n 大小的棋盘，其中 'Q' 代表皇后，'#' 代表空位
|
|
|
let mut state: Vec<Vec<String>> = vec![vec!["#".to_string(); n]; n];
|
|
|
let mut cols = vec![false; n]; // 记录列是否有皇后
|
|
|
let mut diags1 = vec![false; 2 * n - 1]; // 记录主对角线上是否有皇后
|
|
|
let mut diags2 = vec![false; 2 * n - 1]; // 记录次对角线上是否有皇后
|
|
|
let mut res: Vec<Vec<Vec<String>>> = Vec::new();
|
|
|
|
|
|
backtrack(
|
|
|
0,
|
|
|
n,
|
|
|
&mut state,
|
|
|
&mut res,
|
|
|
&mut cols,
|
|
|
&mut diags1,
|
|
|
&mut diags2,
|
|
|
);
|
|
|
|
|
|
res
|
|
|
}
|
|
|
|
|
|
/* Driver Code */
|
|
|
pub fn main() {
|
|
|
let n: usize = 4;
|
|
|
let res = n_queens(n);
|
|
|
|
|
|
println!("输入棋盘长宽为 {n}");
|
|
|
println!("皇后放置方案共有 {} 种", res.len());
|
|
|
for state in res.iter() {
|
|
|
println!("--------------------");
|
|
|
for row in state.iter() {
|
|
|
println!("{:?}", row);
|
|
|
}
|
|
|
}
|
|
|
}
|
