题解 | #蛇形矩阵#有限状态机解蛇形矩阵

看注释解析还是比较简单的，只有4条规则约束

n = int(input())
state = 0
x,y=0,0
position=[[-1 for _ in range(0,n)] for _ in range(0,n)]
trace=[]
cnt = 0
allstate=[[1,-1],[-1,1]]
direction=[[1,0],[0,1]]
# 分别为右上，左下，右，下。
for i in range(0,n**2):
    cnt += 1
    position[x][y] = cnt
    x += allstate[state%2][0]
    y += allstate[state%2][1]
#先向右上移动，如果到墙角了，就回退并且向右走一格
    if (state%2 == 0 or state == 0) and (x>=n or x<0 or y>=n or y<0):
        x -= allstate[state%2][0]
        y -= allstate[state%2][1]
        state += 1
        x += direction[0][0]
        y += direction[0][1]
#向右走一格失败则回退并向下走
        if(x>=n or x<0 or y>=n or y<0):
            x -= direction[0][0]
            y -= direction[0][1]
            x += direction[1][0]
            y += direction[1][1]
            
#先向右上移动，如果到墙角了，就回退并且向右走一格 
    if (state%2 != 0 or state == 1) and (x>=n or x<0 or y>=n or y<0):
        x -= allstate[state%2][0]
        y -= allstate[state%2][1]
        state+=1
        x += direction[1][0]
        y += direction[1][1]
 #向下走一格失败则回退并向右走
        if(x>=n or x<0 or y>=n or y<0):
            x -= direction[1][0]
            y -= direction[1][1]
            x += direction[0][0]
            y += direction[0][1]
        

for i in range(0,n):
    for j in range(0,n):
        print(position[j][i],end=" ")
    print()