因为最近有蓝桥杯考试，但我对于BFS还不是很熟悉。特意这里写文，记录我学习写蓝桥真题 迷宫问题的过程。
package forlanqiao19;
import java.util.Deque;
import java.util.LinkedList;
import java.util.Queue;
public class Main {
public static void main(String[] args) {
String string = "01010101001011001001010110010110100100001000101010"
+"00001000100000101010010000100000001001100110100101"
+"01111011010010001000001101001011100011000000010000"
+"01000000001010100011010000101000001010101011001011"
+"00011111000000101000010010100010100000101100000000"
+"11001000110101000010101100011010011010101011110111"
+"00011011010101001001001010000001000101001110000000"
+"10100000101000100110101010111110011000010000111010"
+"00111000001010100001100010000001000101001100001001"
+"11000110100001110010001001010101010101010001101000"
+"00010000100100000101001010101110100010101010000101"
+"11100100101001001000010000010101010100100100010100"
+"00000010000000101011001111010001100000101010100011"
+"10101010011100001000011000010110011110110100001000"
+"10101010100001101010100101000010100000111011101001"
+"10000000101100010000101100101101001011100000000100" //D -> L -> R -> U
+"10101001000000010100100001000100000100011110101001"
+"00101001010101101001010100011010101101110000110101"
+"11001010000100001100000010100101000001000111000010"
+"00001000110000110101101000000100101001001000011101"
+"10100101000101000000001110110010110101101010100001"
+"00101000010000110101010000100010001001000100010101"
+"10100001000110010001000010101001010101011111010010"
+"00000100101000000110010100101001000001000000000010"
+"11010000001001110111001001000011101001011011101000"
+"00000110100010001000100000001000011101000000110011"
+"10101000101000100010001111100010101001010000001000"
+"10000010100101001010110000000100101010001011101000"
+"00111100001000010000000110111000000001000000001011"
+"10000001100111010111010001000110111010101101111000";
//将上面的一大串字符转为数组中存储，且数组为int型，仅存0 和 1两个值
int[][] moze= new int[30][50];
for (int i = 0 ; i < 30 ; i++) {
for(int j = 0 ; j < 50 ; j++) {
moze[i][j] = string.charAt(i*50+j) - '0'; //
}
}
//要实现bfs的话，需要向队列中不断加入数据，那么这个数据用什么来存储呢？我选择用一个节点来存储每个坐标以及到达该节点的路径。
bfs(moze, 0, 0);
}
//bfs不是递归，无法回溯的。那么如何记录路径呢？我选择在节点中加入路径属性，每次向队列中加入新的节点时，都将该节点的路径初始化为前一个节点的路径+前一个节点到新节点的走法(U/L/R/D);这样的话，当访问到最后一个节点时，输出该节点的路径就是最终答案。
//记录路径长度
static int res;
//记载路径
static int[][] visited = new int[30][50]; //已经走过的就别走了，很关键,否则复杂度将大大增加。
static void bfs(int[][] moze,int row ,int col) {
Deque<Node> deque = new LinkedList<>();
deque.offer(new Node(0,0));
while (!deque.isEmpty()) {
int len = deque.size();
for(int i = 0 ; i < len; i ++) {
Node node = deque.poll();
visited[node.x][node.y] = 1;
if (node.x == 29 && node.y == 49) {
System.out.println(node.path);
return; //找到终点路径，直接退出。
}
if (node.x < 29 && moze[node.x+1][node.y] != 1 && visited[node.x+1][node.y] != 1) { //这里的代码过于冗杂，建议写在方法体内
Node newNode = new Node(node.x+1 , node.y);
deque.offer(newNode);
newNode.path = node.path + 'D';
}
if (node.y > 0 && moze[node.x][node.y-1] != 1 && visited[node.x][node.y-1] != 1) {
Node newNode = new Node(node.x , node.y-1);
deque.offer(newNode);
newNode.path = node.path + 'L';
}
if(node.y < 49 && moze[node.x][node.y+1] != 1 && visited[node.x][node.y+1] != 1) {
Node newNode = new Node(node.x , node.y+1);
deque.offer(newNode);
newNode.path = node.path + 'R';
}
if (node.x > 0 && moze[node.x-1][node.y] != 1 && visited[node.x-1][node.y] != 1) {
Node newNode = new Node(node.x-1 , node.y);
deque.offer(newNode);
newNode.path = node.path + 'U';
}
}
res++; //当每遍历完一层/向前后左右走一次 ，那么我的步数就要加1。
}
System.out.println(res);
}
}
//要实现bfs的话，需要向队列中不断加入数据，那么这个数据用什么来存储呢？我选择用一个节点来存储每个坐标以及到达该节点的路径。
class Node{
int x,y;
String path = "";
public Node() {
// TODO Auto-generated constructor stub
}
Node(int x ,int y ){
this.x = x;
this.y = y;
}
}
