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/*
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* File: n_queens.rs
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* Created Time: 2023-07-15
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* Author: codingonion (coderonion@gmail.com)
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*/
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/* 回溯演算法：n 皇后 */
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fn backtrack(
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row: usize,
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n: usize,
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state: &mut Vec<Vec<String>>,
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res: &mut Vec<Vec<Vec<String>>>,
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cols: &mut [bool],
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diags1: &mut [bool],
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diags2: &mut [bool],
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) {
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// 當放置完所有行時，記錄解
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if row == n {
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res.push(state.clone());
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return;
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}
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// 走訪所有列
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for col in 0..n {
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// 計算該格子對應的主對角線和次對角線
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let diag1 = row + n - 1 - col;
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let diag2 = row + col;
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// 剪枝：不允許該格子所在列、主對角線、次對角線上存在皇后
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if !cols[col] && !diags1[diag1] && !diags2[diag2] {
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// 嘗試：將皇后放置在該格子
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state[row][col] = "Q".into();
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(cols[col], diags1[diag1], diags2[diag2]) = (true, true, true);
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// 放置下一行
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backtrack(row + 1, n, state, res, cols, diags1, diags2);
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// 回退：將該格子恢復為空位
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state[row][col] = "#".into();
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(cols[col], diags1[diag1], diags2[diag2]) = (false, false, false);
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}
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}
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}
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/* 求解 n 皇后 */
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fn n_queens(n: usize) -> Vec<Vec<Vec<String>>> {
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// 初始化 n*n 大小的棋盤，其中 'Q' 代表皇后，'#' 代表空位
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let mut state: Vec<Vec<String>> = vec![vec!["#".to_string(); n]; n];
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let mut cols = vec![false; n]; // 記錄列是否有皇后
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let mut diags1 = vec![false; 2 * n - 1]; // 記錄主對角線上是否有皇后
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let mut diags2 = vec![false; 2 * n - 1]; // 記錄次對角線上是否有皇后
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let mut res: Vec<Vec<Vec<String>>> = Vec::new();
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backtrack(
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0,
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n,
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&mut state,
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&mut res,
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&mut cols,
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&mut diags1,
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&mut diags2,
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);
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res
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}
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/* Driver Code */
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pub fn main() {
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let n: usize = 4;
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let res = n_queens(n);
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println!("輸入棋盤長寬為 {n}");
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println!("皇后放置方案共有 {} 種", res.len());
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for state in res.iter() {
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println!("--------------------");
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for row in state.iter() {
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println!("{:?}", row);
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}
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}
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}
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