题目：
班级组织传球活动，男女同学随机排成m行n列队伍，第一列中的任意一个男同学都可以作为传球的起点，要求最终将球传到最后一列的任意一个男同学手里，求所有能够完成任务的传球路线中的最优路线(传球次数最少的路线)的传球次数。

传球规则：
1.男同学只能将球传给男同学，不能传给女同学。
2.球只能传给身边前后左右相邻的同学。
3.如果游戏不能完成，返回-1。

说明：
1.传球次数最少的路线为最优路线。
2.最优路线可能不唯一，不同最优路线都为最少传球次数。

解答要求：
时间限制:C/C++100ms其他语言: 200ms内存限制: C/C++256MB,其他语言: 512MB

输入：
班级同学随机排成的m行n列队伍，1代表男同学，0代表女同学。
输入第一行包含两个用空格分开的整数m[1,30]和n [1,30]，表示m行n列的队伍;
接下来是m行每行包含n个用空格分开的整数1或0。

输出：
最优路线的传球次数（最少传球次数）

样例
输入：
4 4
1 1 1 0
1 1 1 0
0 0 1 0
0 1 1 1
输出：
5

//DFS
import java.util.Scanner;public class Huawei {private static int minPasses = Integer.MAX_VALUE;private static int[] dx = {-1, 1, 0, 0};private static int[] dy = {0, 0, -1, 1};public static void main(String[] args) {Scanner scanner = new Scanner(System.in);int m = scanner.nextInt();int n = scanner.nextInt();int[][] grid = new int[m][n];for (int i = 0; i < m; i++) {for (int j = 0; j < n; j++) {grid[i][j] = scanner.nextInt();}}for (int i = 0; i < m; i++) {if (grid[i][0] == 1) {dfs(i, 0, 0, m, n, grid);}}if (minPasses == Integer.MAX_VALUE) {System.out.println(-1);} else {System.out.println(minPasses);}}private static void dfs(int x, int y, int passes, int m, int n, int[][] grid) {if (y == n - 1) {minPasses = Math.min(minPasses, passes);return;}for (int i = 0; i < 4; i++) {int nx = x + dx[i];int ny = y + dy[i];if (nx >= 0 && nx < m && ny >= 0 && ny < n && grid[nx][ny] == 1) {grid[x][y] = 0;dfs(nx, ny, passes + 1, m, n, grid);grid[x][y] = 1;}}}
}

//BFS，代码来源万诺coding
import java.io.*;
import java.util.Deque;
import java.util.LinkedList;
import java.util.StringTokenizer;public class Main {private static  class FastScanner {BufferedReader br;StringTokenizer st;public FastScanner(InputStream stream){br =  new BufferedReader(new InputStreamReader(stream), 32768);st = null;}String next() {while (st == null ||  !st.hasMoreTokens())try {st=new StringTokenizer(br.readLine());} catch (IOException e) {e.printStackTrace();}return st.nextToken();}int nextInt() {return Integer.parseInt(next());}}public static void main(String[] args) {new Main().solve();}int m,n;int[][] grid;void solve() {PrintWriter pwin = new PrintWriter(new OutputStreamWriter(System.out));FastScanner fsc = new FastScanner(System.in);m = fsc.nextInt();;n = fsc.nextInt();grid = new int[m][n];for (int i = 0; i < m; i++) {for (int j = 0; j < n; j++) {grid[i][j] = fsc.nextInt();}}int[][] dirs = {{0,1},{1,0},{0,-1},{-1,0}};int res = 10000001;for (int i = 0; i < m; i++) {if (grid[i][0] == 1) {Deque<int[]> dq = new LinkedList<>();dq.add(new int[]{0,i,0});boolean[][] used = new boolean[m][n];used[i][0] = true;while (!dq.isEmpty()) {int[] a = dq.poll();int d = a[0], x = a[1], y = a[2];if (y == n-1) {res = Math.min(res, d);}for (int[] dir : dirs) {int nx = x + dir[0],  ny = y + dir[1];if (nx < 0 || ny < 0 || nx >= m || ny >= n || used[nx][ny] || grid[nx][ny] != 1) continue;used[nx][ny] = true;dq.add(new int[]{d+1, nx,ny});}}}}if (res != 10000001) pwin.println(res);else pwin.println(-1);pwin.flush();}
}