hello-algo/codes/rust/chapter_backtracking/n_queens.rs
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idiomatic rust (#1485)
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2024-08-23 02:33:47 +08:00

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/*
* 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 {}", res.len());
for state in res.iter() {
println!("--------------------");
for row in state.iter() {
println!("{:?}", row);
}
}
}