題目內容為一個棋盤，長寬自己輸入，棋盤狀態自己輸入（.表示無棋子，o表示有棋子），可以全部進行上移、下移、左移、右移操作，遇到邊界棋子丟失，計算出最短步數使棋盤上僅剩下k個棋子。

import java.util.*;
public class BFS {

    public static int n = 0;
    public static int m = 0;

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String s = sc.nextLine();
        String[] strs = s.split(" ");
        n = Integer.parseInt(strs[0]);
        m = Integer.parseInt(strs[1]);
        char[][] ch = new char[n][m];
        for(int i = 0; i < n; i++){
            String str = sc.nextLine();
            ch[i] = str.toCharArray();
        }

        int k = sc.nextInt();
        System.out.print(bfs(ch,k));
    }

    public static int bfs(char[][] ch, int k){
        int currentK = count(ch);
        if(currentK < k)
            return -1;
        if(currentK == k)
            return 0;

        Queue<char[][]> queue = new ArrayDeque<>();
        int step = 0;
        queue.offer(ch);
        Queue<Integer> q = new ArrayDeque<>();
        Set<char[][]> set = new HashSet<>();
        set.add(ch);
        q.add(step);
        while(!queue.isEmpty()){
            char[][] current = queue.poll();
            int num = count(current);
            step = q.poll();
            if(num == k){
                return step;
            }
            //expand
            step++;
            char[][] upChar = up(current);
            currentK = count(upChar);
            if(currentK >= k && !set.contains(upChar)){
                queue.offer(upChar);
                q.offer(step);
                set.add(upChar);
            }

            char[][] downChar = down(current);
            currentK = count(downChar);
            if(currentK >= k && !set.contains(downChar)){
                queue.offer(downChar);
                q.offer(step);
                set.add(downChar);
            }

            char[][] leftChar = left(current);
            currentK = count(leftChar);
            if(currentK >= k && !set.contains(leftChar)){
                queue.offer(leftChar);
                q.offer(step);
                set.add(leftChar);
            }

            char[][] rightChar = right(current);
            currentK = count(rightChar);
            if(currentK >= k && !set.contains(rightChar)){
                queue.offer(rightChar);
                q.offer(step);
                set.add(rightChar);
            }
        }
        return -1;

    }

    public static char[][] up(char[][] ch){
        char[] tmp = {'.','.','.','.'};
        char[][] res = new char[n][m];
        for(int i = 0; i < n-1; i++){
            res[i] = ch[i+1];
        }
        res[n-1] = tmp;
        return res;
    }

    public static char[][] down(char[][] ch){
        char[] tmp = {'.','.','.','.'};
        char[][] res = new char[n][m];
        for(int i = n-1; i > 0; i--){
            res[i] = ch[i-1];
        }
        res[0] = tmp;
        return res;
    }

    public static char[][] left(char[][] ch){
        char[][] res = new char[n][m];
        for(int j = 0; j < m-1; j++){
            for(int i = 0; i < n; i++){
                res[i][j] = ch[i][j+1];
            }
        }
        for(int i = 0; i < n; i++){
            res[i][m-1] = '.';
        }
        return res;
    }

    public static char[][] right(char[][] ch){
        char[] tmp = {'.','.','.','.'};
        char[][] res = new char[n][m];
        for(int j = m-1; j > 0; j--){
            for(int i = 0; i < n; i++){
                res[i][j] = ch[i][j-1];
            }
        }
        for(int i = 0; i < n; i++){
            res[i][0] = '.';
        }
        return res;
    }

    public static int count(char[][] ch){
        int cnt = 0;
        for(int i = 0; i < n; i++){
            for(int j = 0; j < m; j++){
                if(ch[i][j] == 'o')
                    cnt++;
            }
        }
        return cnt;
    }
}