BFS_2:迷宫最短路径

BFS_2:迷宫最短路径

输入样例：

5 5 
1 0 1 1 1 
0 0 1 1 0
1 0 1 1 1
0 0 1 1 1
0 0 0 1 1 

5
(5,3)->(5,2)->(4,2)->(3,2)->(2,2)

问题分析：

该题为经典题，典型的BFS搜索最短路径，比较要注意的是pre成员属性，用于记录当前节点的前一个节点，从而寻找最短的路径。

二、使用步骤

import java.util.*;
class Node{
    public int x;
    public int y;
    public int dis;
    public Node pre;//用来记录路径
    public Node(int x, int y) {
        this.x = x;
        this.y = y;
    }
    public Node(int x, int y, int dis) {
        this.x = x;
        this.y = y;
        this.dis = dis;
    }
}
public class BFS_2 {
    static int N = 40;
    static int n,m;
    static int[][] mp = new int[N][N];
    static int[][] vis = new int[N][N];
    static int[] xx = {0 ,0 , -1 , 1};
    static int[] yy = {-1 ,1 ,0 , 0 };
    static Node end;
    static Node start;
    static List<Node> path = new ArrayList<>();
    static Queue<Node> queue = new LinkedList<>();
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        n = scanner.nextInt();
        m = scanner.nextInt();
        start = new Node(1,2,0) ;
        end = new Node(5,3) ;
        for (int i = 1; i <= n ; i++) {
            for (int j = 1; j <= m ; j++) {
                mp[i][j]  = scanner.nextInt();
            }
        }
         bfs(start);
    }

    private static void bfs(Node node) {
        //将队首元素压入队列
        StringBuilder s = new StringBuilder();
        queue.offer(node);
        vis[node.x][node.y] = 1;
        //开始进行搜索
        while (!queue.isEmpty()){
            Node cur  = queue.peek();
            //如果当前的点是终点的话，则退出
            if(cur.x==end.x&&cur.y== end.y) {
                System.out.println(cur.dis);
                System.out.print("("+cur.x+","+cur.y+")->");
                    for (int j = path.size()-1; j >= 0 ; j--) {
                        if(path.get(j) == cur.pre){
                            s.append("(").append(path.get(j).x).append(",").append(path.get(j).y).append(")->");
                            cur = path.get(j);
                        }
                    }
                    s.delete(s.length()-2,s.length());//处理尾巴
                    System.out.println(s);
            }
            for (int i = 0; i < 4; i++) {
                int dx = cur.x + xx[i];
                int dy = cur.y + yy[i];
                //判断该点是否可以走
                if(dx>=1&&dx<=n&&dy>=1&&dy<=m&&mp[dx][dy]==0&&vis[dx][dy]==0){
                    Node temp = new Node(dx,dy);
                    vis[temp.x][temp.y]= 1 ;
                    temp.dis = cur.dis + 1;
                    temp.pre = cur;
                    path.add(temp);
                    //入队
                    queue.offer(temp);
                }
            }
            //当队首元素的所有方向都找完之后，出队列
            queue.poll();
        }
    }
}