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黑白棋（Othello, ACM/ICPC World Finals 1992, UVa220）rust解法

news 2025/8/13 7:31:46

你的任务是模拟黑白棋游戏的进程。黑白棋的规则为：黑白双方轮流放棋子，每次必须让新放的棋子“夹住”至少一枚对方棋子，然后把所有被新放棋子“夹住”的对方棋子替换成己方棋子。一段连续（横、竖或者斜向）的同色棋子被“夹住”的条件是两端都是对方棋子（不能是空位）。如图4-6（a）所示，白棋有6个合法操作，分别为(2,3),(3,3),(3,5),(6,2),(7,3),(7,4)。选择在(7,3)放白棋后变成如图4-6（b）所示效果（注意有竖向和斜向的共两枚黑棋变白）。注意(4,6)的黑色棋子虽然被夹住，但不是被新放的棋子夹住，因此不变白。
在这里插入图片描述输入一个8*8的棋盘以及当前下一次操作的游戏者，处理3种指令：

L指令打印所有合法操作，按照从上到下，从左到右的顺序排列（没有合法操作时输出No legal move）。
Mrc指令放一枚棋子在(r,c)。如果当前游戏者没有合法操作，则是先切换游戏者再操作。输入保证这个操作是合法的。输出操作完毕后黑白方的棋子总数。
Q指令退出游戏，并打印当前棋盘（格式同输入）。

样例：
输入

--------
--------
--------
---WB---
---BW---
--------
--------
--------
W
L
M35
L
Q

输出

[(3, 5), (4, 6), (5, 3), (6, 4)]
W is 4. B is 1
[(3, 4), (3, 6), (5, 6)]
--------
--------
----W---
---WW---
---BW---
--------
--------
--------

解法：

use std::io;enum Cmd {Print,Move(usize, usize),Quit,
}
fn main() {let mut grid: Vec<Vec<char>> = vec![];for _i in 0..8 {let mut buf = String::new();io::stdin().read_line(&mut buf).unwrap();grid.push(buf.trim().chars().collect());}let mut buf = String::new();io::stdin().read_line(&mut buf).unwrap();let mut curplayer = buf.trim().chars().nth(0).unwrap();let mut cmds: Vec<Cmd> = vec![];loop {let mut buf = String::new();io::stdin().read_line(&mut buf).unwrap();buf = buf.trim().to_string();if buf == "L" {cmds.push(Cmd::Print);} else if buf == "Q" {cmds.push(Cmd::Quit);break;} else {let i = buf.chars().nth(1).unwrap().to_digit(10).unwrap();let j = buf.chars().nth(2).unwrap().to_digit(10).unwrap();cmds.push(Cmd::Move(i as usize, j as usize));}}for i in cmds {match i {Cmd::Print => printmoves(&grid, curplayer),Cmd::Move(x, y) => fangzi(&mut grid, &mut curplayer, (x, y)),Cmd::Quit => {printgrid(&grid);break;}}}
}fn fangzi(grid: &mut Vec<Vec<char>>, curp: &mut char, pos: (usize, usize)) {let allmoves = getmoves(grid, *curp);if allmoves.is_empty() {*curp = oposite(*curp);}let newpos = (pos.0 - 1, pos.1 - 1);grid[newpos.0][newpos.1] = *curp;let runs = [(0, -1),(0, 1),(-1, 0),(1, 0),(-1, -1),(1, 1),(-1, 1),(1, -1),];for d in runs {if judge(grid, *curp, newpos, d) {change(grid, *curp, newpos, d);}}let nums = getnums(grid);println!("W is {}. B is {}", nums.0, nums.1);*curp = oposite(*curp);
}fn getnums(grid: &Vec<Vec<char>>) -> (u32, u32) {let mut nums = (0, 0);for i in 0..8 {for j in 0..8 {if grid[i][j] == 'W' {nums.0 += 1;} else if grid[i][j] == 'B' {nums.1 += 1;}}}return nums;
}
fn oposite(p: char) -> char {if p == 'W' {'B'} else {'W'}
}
fn printgrid(grid: &Vec<Vec<char>>) {for line in grid.iter() {println!("{}", line.iter().collect::<String>());}
}
fn getmoves(grid: &Vec<Vec<char>>, curp: char) -> Vec<(usize, usize)> {let mut allmoves: Vec<(usize, usize)> = vec![];let runs = [(0, -1),(0, 1),(-1, 0),(1, 0),(-1, -1),(1, 1),(-1, 1),(1, -1),];for i in 0..8 {for j in 0..8 {//println!("i,j: {},{}", i, j);if grid[i][j] != '-' {continue;}for d in runs {//检查八个方向if judge(grid, curp, (i, j), d) {allmoves.push((i + 1, j + 1));}}}}return allmoves;
}
fn judge(grid: &Vec<Vec<char>>, curp: char, pos: (usize, usize), run: (i32, i32)) -> bool {let mut x = pos.0;let mut y = pos.1;let mut bjiazhu = false;while x > 0 && x < 7 && y > 0 && y < 7 {x = (x as i32 + run.0) as usize;y = (y as i32 + run.1) as usize;if grid[x][y] == '-'{break;}if grid[x][y] == oposite(curp) {bjiazhu = true;}else if bjiazhu {return true;}else {break;}}return false;
}
fn change(grid: &mut Vec<Vec<char>>, curp: char, pos: (usize, usize), run: (i32, i32)) {let mut x = pos.0;let mut y = pos.1;while x > 0 && x < 7 && y > 0 && y < 7 {x = (x as i32 + run.0) as usize;y = (y as i32 + run.1) as usize;if grid[x][y] == oposite(curp) {grid[x][y] = curp;} else {return;}}
}
fn printmoves(grid: &Vec<Vec<char>>, curp: char) {let allmoves = getmoves(grid, curp);if allmoves.is_empty() {println!("No legal move");} else {println!("{:?}", allmoves);}
}