Nothing is impossible as long as you try your best,Because everything is possible!

一个很屑的代码(似乎有点抽象)

#未完成的、基于BFS画树的Python代码。
import turtle
#import time
#p=turtle.Turtle()
#vis = []
def move(x,y):#自定义
    turtle.penup()
    turtle.goto(x,y)
    turtle.pendown()
    #turtle.dot(3)
queue=[]#广度优先搜索用到队列,这里用于存储坐标
#deep =int(input("请输入树的深度:")) #限制迭代深度
#length = int(input("请输入树枝的长度:")) 
def draw(angle):
    turtle.setheading(angle)
    turtle.forward(length)
    queue.append(turtle.pos())
    turtle.backward(length)
deep = 6
length = 30
init()
cnt = 0
angle = 30 
queue.append((-100,0))
i=0
while cnt <= 2**deep:#BFS
    u=len(queue)
    p = queue[0]
    queue.pop(0)#出队
    move(p[0],p[1])
    x,y=turtle.pos()
    angle = (angle+2)%90
    draw(angle)
    draw(360-angle)
    i+=1
    cnt += 1
turtle.done()

还有更屑的：

#include <bits/stdc++.h>
using namespace std;
int a[15][15],val[15][15],n,m,T,pos;
char x;
struct Node{
	int x,y;
}e[15];
queue <Node> Q;
bool search(){
    for (int i=1;i<=n;i++){
        for (int j=1;j<=m;j++){

        }
    }
    return false;
}
int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	cin>>T;
	while (T--){
		cin>>n>>m;
        pos = 0;
		for (int i=1;i<=n;i++)
			for (int j=1;j<=m;j++){
				cin>>x;
				if (x == '*') a[i][j] = 1;
				else if (x >= 48 && x<= 57) val[i][j] = x - 48;
				else e[++pos].x=i,e[++pos].y=j;
			}
        const int l=pos;
		bitset<l> bs;
        bool flag = false;
		for (int i=0;i<(1<<l);i++){
		    bs = i;
		    for (int j=0;j<l;j++)
                a[e[bs[j]].x][e[bs[j]].y] = bs[j];
            if (search()){
                flag =true;
                break;
            }
		}
        (!flag) ? cout<<"YES"<<"\n" : cout<<"NO"<<"\n";
		memset(a,0,sizeof(a));	
		memset(val,0,sizeof(val));
	} 
	return 0;
}