Labyrinth path recording bfs +

Given Map 5 * 5, 1 wall, 0 for space, walked the lower left corner from the top left corner

Is to go all the paths are recorded, but the
search recursive and non-recursive form

This code uses a non-recursive form
bfs + recording path

For num [i] .pre = head his understanding there are many routes, each point is a point came from, but there is always a path to reach the terminal is, therefore, just we need to get to a head end you can follow the path marked back

#include <iostream>
#include <cstdio>
using namespace std;
char a[10][10];
int dir[4][2]={1,0,-1,0,0,1,0,-1};
struct node{
    int x,y;
    int pre;
}num[1200];
void print(int n){//输出路径，当xx与yy到达终点的时候，就把head传入，
    if(num[n].pre != -1){  
        print(num[n].pre);  
        printf("(%d, %d)\n",num[n].x,num[n].y);
    } 
}
void bfs(){
    int head=0;
    int rear=0;
    num[rear].x=0,num[rear].y=0,num[rear++].pre=-1;//初始路径记录
    while(head<rear){
        for(int i=0;i<4;i++){
            int xx=num[head].x+dir[i][0];
            int yy=num[head].y+dir[i][1];
            if(xx>=5||xx<0||yy>=5||yy<0||a[xx][yy]==1)continue;
            a[xx][yy]=1;//标记走过了
            num[rear].x=xx;
            num[rear].y=yy;
            num[rear].pre=head;
            rear++;//rear的值一只增加，只要一个点的4个方向存在空地
            if(xx==4&&yy==4)print(head);
        }
        head++;

    }
}
int main(){
    for(int i=0;i<5;i++){
        for(int j=0;j<5;j++){
            scanf("%d",&a[i][j]);
        }
    }  
    printf("(0, 0)\n");
    bfs();
    printf("(4, 4)\n");
    
    return 0;
}