迷宫的最短路径--BFS

import java.util.LinkedList;
import java.util.Queue;
import java.util.Scanner;
class P {
    int x;
    int y;
    public P(int a, int b) {
        this.x = a;
        this.y = b;
    }
}
public class Main {
    static final int INF = -1;
    static char[][] arr;
    static int[][] dis;
    static int[] tempx = { 1, 0, -1, 0 };
    static int[] tempy = { 0, 1, 0, -1 };
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        int M = sc.nextInt();
        int sx = 0, sy = 0, gx = 0, gy = 0;
        arr = new char[N][M];
        dis = new int[N][M];
        for (int i = 0; i < N; i++) {                           // 输入 以及得到sx sy 的地址
            String str = sc.next();
            arr[i] = str.toCharArray();
            for (int x = 0; x < str.length(); x++) {
                dis[i][x] = INF;                                            // 把所有位置初始化为INF 初始化
                if (str.charAt(x) == 'S') { 
                    sx = i;
                    sy = x;
                }
                if (str.charAt(x) == 'G') {
                    gx = i;
                    gy = x;
                }
            }
        }
        dis[sx][sy] = 0;                                                                //将起点设置为 0
        Queue<P> que = new LinkedList<P>();                 //创建一个队列
        que.offer(new P(sx, sy));                                           //直接扔到队列里面
//      System.out.println(sx + " " + sy + " " + gx + " " + gy);
        while (que.size() > 0) {
            P p = que.poll();                                                           //poll队列的头
            if (p.x == gx && p.y == gy)                                     //到达终点跳出来
                break;
 
            for (int i = 0; i < 4; i++) {                                          //四个方向
                int nx = p.x + tempx[i];
                int ny = p.y + tempy[i];
 
                if (nx >= 0 && nx < N && ny >= 0 && ny < M && dis[nx][ny] == INF && arr[nx][ny] != '#') {
                    que.offer(new P(nx, ny));
                    dis[nx][ny] = dis[p.x][p.y] + 1;
                }
            }
        }
        System.out.println(dis[gx][gy]);
    }
}