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

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。

解法

use std::io;

fn main() {
    let mut grid: Vec> = 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] == '0' {
                kong = (i, j);
                break;
            }
        }
    }
    for i in cmds.chars() {
        match i {
            'a' => {
                if kong.0 < 1 {
                    panic!("This puzzle has no final configuration");
                }
                let c = grid[kong.0 - 1][kong.1];
                grid[kong.0][kong.1] = c;
                grid[kong.0 - 1][kong.1] = '0';
                kong = (kong.0 - 1, kong.1);
            }
            'b' => {
                if kong.0 >= 4 {
                    panic!("This puzzle has no final configuration");
                }
                let c = grid[kong.0 + 1][kong.1];
                grid[kong.0][kong.1] = c;
                grid[kong.0 + 1][kong.1] = '0';
                kong = (kong.0 + 1, kong.1);
            }
            'l' => {
                if kong.1 < 1 {
                    panic!("This puzzle has no final configuration");
                }
                let c = grid[kong.0][kong.1 - 1];
                grid[kong.0][kong.1] = c;
                grid[kong.0][kong.1 - 1] = '0';
                kong = (kong.0, kong.1 - 1);
            }
            'r' => {
                if kong.1 >= 4 {
                    panic!("This puzzle has no final configuration");
                }
                let c = grid[kong.0][kong.1 + 1];
                grid[kong.0][kong.1] = c;
                grid[kong.0][kong.1 + 1] = '0';
                kong = (kong.0, kong.1 + 1);
            }
            '0' => {
                break;
            }
            _ => {
                panic!("This puzzle has no final configuration");
            }
        }
    }
    println!("{:#?}", grid);
}