Backtracking maze problem

to flash back

Examples of data structures

Ask to get out of the maze , The maze is as follows
0,0,0,0,0,0,0,0,0,0
0,1,1,0,1,1,1,0,1,0
0,1,1,0,1,1,1,0,1,0
0,1,1,1,1,0,0,1,1,0
0,1,0,0,0,1,1,1,1,0
0,1,1,1,0,1,1,1,1,0
0,1,0,1,1,1,0,1,1,0
0,1,0,0,0,1,0,0,1,0
0,0,1,1,1,1,1,1,1,0
0,0,0,0,0,0,0,0,0,0
here 0 Representative wall ,1 Represents the road that can be taken .
It can be used DFS and BFS Write , But I won't hahaha .
Here is the code

#include<iostream>
using namespace std;

#define M 100

#define MAX 0x3f3f3f3f
typedef long long ll;
#define bug(a) cout<<endl<<"*"<<a<<endl;

int arr[10][10] = {
     0,0,0,0,0,0,0,0,0,0,//0
				    0,1,1,0,1,1,1,0,1,0,//1
				    0,1,1,0,1,1,1,0,1,0,//2
				    0,1,1,1,1,0,0,1,1,0,//3
				    0,1,0,0,0,1,1,1,1,0,//4
				    0,1,1,1,0,1,1,1,1,0,//5
				    0,1,0,1,1,1,0,1,1,0,//6
				    0,1,0,0,0,1,0,0,1,0,//7
				    0,0,1,1,1,1,1,1,1,0,//8
				    0,0,0,0,0,0,0,0,0,0 };//9
int dic[4][2]={
    1,0,0,1,-1,0,0,-1};// Coordinates of movement in four directions 
int brr[10][10];
typedef struct node1
{
    
	int h, l;
}qo;
typedef struct node
{
    
	int ord;
	qo* seat;// coordinate 
	qo* base;// Bottom 
	int di;// Direction 
}stack;
void newstack(stack& s) {
    
	s.base = new qo[M];// Both need to create space 
	s.seat = new qo[M];// perhaps s.seat Do not create space s.seat=s.base It's OK 
	if (!s.base) exit(0);
	if (!s.seat) exit(0);
}
void push(stack& s, qo& b) {
    // Put the value on the stack 
	*(s.seat++) = b;
	s.ord++;
}
void pop(stack& s) {
    // Pop up elements , But it doesn't return a value 
	if (s.seat == s.base)exit(0);
	s.ord--;
	*(s.seat--);
}
void popp(stack& s, qo& a2) {
    // Pop up the element and bring it out 
	if (s.ord==0)exit(0);
	s.ord--;
	*(s.seat--);
	a2 = *s.seat;
}
int main()
{
    
	int ans1 = 8,flag=0;//8 Because the exit is in the coordinates 8,8 The location of 
	stack s;
	s.ord = 0;// Remember to assign values to this tag , Otherwise, no output 
	qo a1;
	newstack(s);// Create space , Don't forget 
	a1.h = 1;a1.l = 1;
	brr[1][1] = 1;// The road signs we have passed 1, The starting position is also marked 
	push(s, a1);
	while (1) {
    
		if (a1.h == ans1 && a1.l == ans1) {
     flag = 1; break; }// Export order flag=0
		if (s.ord==0)break;// An empty stack means there is no way to the exit , Just jump out 
		if (arr[a1.h + dic[0][0]][a1.l + dic[0][1]] != 0&&brr[a1.h + dic[0][0]][a1.l + dic[0][1]]==0) {
    
			a1.h += dic[0][0];
			a1.l += dic[0][1];
			brr[a1.h][a1.l] = 1;
			push(s, a1);
		}
		else {
    
			int i;
			for (i = 1; i <= 3; i++) {
    
				if (arr[a1.h + dic[i][0]][a1.l + dic[i][1]] != 0 && brr[a1.h + dic[i][0]][a1.l + dic[i][1]] == 0) {
    
					a1.h += dic[i][0];
					a1.l += dic[i][1];
					brr[a1.h][a1.l] = 1;
					push(s, a1);
					break;
				}
			}
			if (i > 3) {
    // It means 4 Neither direction satisfies the condition , To pop up a , Equivalent to backtracking 
				pop(s);
				a1 = *--s.seat;
				*(s.seat++) = a1;// This means that a value will pop up for a2 however a2 Again 
				brr[a1.h][a1.l] = 1;// Giving it back is equivalent to not playing it out but giving it a value 
			}
		}
	}
	if (!flag)cout << "eror" << endl;// have no way out 
	else if (flag) {
    // There is a way 
		qo a2;
		memset(brr, 0, sizeof(brr));// Clear the road you have walked before 0
		int k = s.ord;
		for (int j = 1; j <= k; j++) {
    
			popp(s, a2);
			brr[a2.h][a2.l] = 1; A road sign that will lead to the end 1
		}
		for (int i = 0; i < 10; i++) {
    
			for (int j = 0; j < 10; j++) {
    
				if (brr[i][j] == 1)cout << "@" << ' ';// The way out is used @ Mark ;
				else cout << "#" << ' ';
			}
			cout << endl;
		}
	}
	return 0;
}