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* Revised the book * Update the book with the second revised edition * Revise base on the manuscript of the first edition
72 lines
2.4 KiB
Rust
72 lines
2.4 KiB
Rust
/*
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* File: n_queens.rs
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* Created Time: 2023-07-15
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* Author: sjinzh (sjinzh@gmail.com)
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*/
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/* 回溯算法：N 皇后 */
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fn backtrack(row: usize, n: usize, state: &mut Vec<Vec<String>>, res: &mut Vec<Vec<Vec<String>>>,
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cols: &mut [bool], diags1: &mut [bool], diags2: &mut [bool]) {
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// 当放置完所有行时，记录解
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if row == n {
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let mut copy_state: Vec<Vec<String>> = Vec::new();
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for s_row in state.clone() {
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copy_state.push(s_row);
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}
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res.push(copy_state);
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return;
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}
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// 遍历所有列
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for col in 0..n {
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// 计算该格子对应的主对角线和次对角线
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let diag1 = row + n - 1 - col;
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let diag2 = row + col;
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// 剪枝：不允许该格子所在列、主对角线、次对角线上存在皇后
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if !cols[col] && !diags1[diag1] && !diags2[diag2] {
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// 尝试：将皇后放置在该格子
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state.get_mut(row).unwrap()[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：将该格子恢复为空位
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state.get_mut(row).unwrap()[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 大小的棋盘，其中 'Q' 代表皇后，'#' 代表空位
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let mut state: Vec<Vec<String>> = Vec::new();
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for _ in 0..n {
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let mut row: Vec<String> = Vec::new();
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for _ in 0..n {
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row.push("#".into());
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}
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state.push(row);
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}
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let mut cols = vec![false; n]; // 记录列是否有皇后
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let mut diags1 = vec![false; 2 * n - 1]; // 记录主对角线上是否有皇后
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let mut diags2 = vec![false; 2 * n - 1]; // 记录次对角线上是否有皇后
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let mut res: Vec<Vec<Vec<String>>> = Vec::new();
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backtrack(0, n, &mut state, &mut res, &mut cols, &mut diags1, &mut diags2);
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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!("输入棋盘长宽为 {n}");
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println!("皇后放置方案共有 {} 种", 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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