谜题（Puzzle, ACM/ICPC World Finals 1993, UVa227）rust解法

有一个5*5的网格，其中恰好有一个格子是空的，其他格子各有一个字母。一共有4种指令：A, B, L, R，分别表示把空格上、下、左、右的相邻字母移到空格中。输入初始网格和指令序列（以数字0结束），输出指令执行完毕后的网格。如果有非法指令，应输出“This puzzle has no final configuration.”，例如，图3-5中执行ARRBBL0后，效果如图3-6所示。

样例：
输入

TRGSJ
XDOKI
M VLN
WPABE
UQHCF
ARRBBL0
1
2
3
4
5
6

输出

TRGSJ
XOKLI
MDVBN
WP AE
UQHCF
1
2
3
4
5

解法：

use std::{collections::HashMap, io};

fn main() {
    let mut grid: Vec<Vec<char>> = vec![];
    for _i in 0..5 {
        let mut buf = String::new();
        io::stdin().read_line(&mut buf).unwrap();
        let cs = buf.trim().chars().collect();
        grid.push(cs);
    }
    //println!("{:#?}", grid);

    let mut buf = String::new();
    io::stdin().read_line(&mut buf).unwrap();
    let cmds = buf.trim();
    //println!("{}", cmds);

    let mut kong = (0, 0);
    for i in 0..5 {
        for j in 0..5 {
            if grid[i][j] == ' ' {
                kong = (i, j);
                break;
            }
        }
    }
    let run_directions =
        HashMap::from([('A', (-1, 0)), ('B', (1, 0)), ('L', (0, -1)), ('R', (0, 1))]);
    for i in cmds.chars() {
        if kong.0 < 1 || kong.0 >= 4 || kong.1 < 1 || kong.1 >= 4 {
            panic!("This puzzle has no final configuration");
        }
        if i == '0' {
            break;
        }
        let d = run_directions.get(&i).unwrap();
        let x = (kong.0 as i32 + d.0) as usize;
        let y = (kong.1 as i32 + d.1) as usize;
        let c = grid[x][y];
        grid[kong.0][kong.1] = c;
        grid[x][y] = ' ';
        kong = (x, y);
    }
    //println!("{:#?}", grid);
    for v in grid.iter(){
        println!("{}", v.iter().collect::<String>());
    }
}
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