Regarding the understanding of the N Queens problem and the backtracking method, I personally recommend the following video: Algorithms and data structures, backtracking to solve the eight queens, the most classic recursive problem_哔哩哔哩_bilibili Eight Queens problem is the most classic in computer science One of the problems, which was posed by international chess player Max Bethel in 1848. The eight queens problem is to place eight queens on an 8×8 chessboard so that they cannot attack each other. Any two queens cannot be in the same row, column or slant. Find out how many queens there are Placement methods, specific placement methods for each placement method. 
https://www.bilibili.com/video/BV1ZK411K7A8?share_source=copy_web
The code in the video is written in C++. I have more personal experience with embedded, so I am more accustomed to using C.
The C language code is as follows:
/* N皇后问题 */
#include <stdio.h>
#include <stdlib.h>
#define N 100 //宏定义最大的N
int attack[N][N];
char queen[N][N];
int cnt;
void init(int n);//初始化attack和queen数组
void print(int n);//打印符合要求的摆法
void print_ack(int n);//测试update_ack函数编写是否正确
void copy(int object[N][N],int target[N][N],int n);//将对象数组保存给目标数组
void update_ack(int x,int y,int n);//摆放queen后更新attack数组，将进queen攻路线上的值标为1
void backtrack(int row,int n);//回溯法 递归调用
int main(int argc, char const *argv[])
{
int n;
puts("Please input the N of queen:");
scanf("%d",&n);//计算nXn的皇后问题
init(n);//初始化attack和queen数组
backtrack(0,n);//从第0行开始摆放queen
printf("The number of required placement method =%d\n",cnt);
system("pause");
return 0;
}
void backtrack(int row,int n)
{
if(row==n)//递归执行到摆放完所有行的结束条件
{
cnt++;
print(n);
return ;
}
int col;
for(col=0;col<n;col++)
{
if(attack[row][col]==0)//判断该位置是能够放置queen
{
int tem[N][N];//保存当前的attack数组内容，便于回溯
copy(attack,tem,n);//遍历交换
queen[row][col]='Q';//该位置放置queen
update_ack(row,col,n);//放置完后将棋盘上queen的进攻路线全部标为1
backtrack(row+1,n);//行数+1，递归回到函数头部重复判断
//递归结束才会执行下面的语句
copy(tem,attack,n);//将之前保存的temp返还给attack
queen[row][col]='.';//将标记的Q改回.
}
}
}
void init(int n)//初始化attack和queen数组
{
int i,j;
for(i=0;i<n;i++)//对二维数组遍历
for(j=0;j<n;j++)
{
attack[i][j]=0;
queen[i][j]='.';
}
}
void print(int n)//打印符合要求的摆法
{
int i,j;
printf("cnt=%d\n",cnt);
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
printf("%c\t",queen[i][j]);
}
printf("\n");
}
printf("\n");
}
void print_ack(int n)//测试update_ack函数编写是否正确
{
int i,j;
for(i=0;i<n;i++)//对二维数组遍历
{
for(j=0;j<n;j++)
{
printf("%d\t",attack[i][j]);
}
printf("\n");//每打印出n个换行
}
}
void copy(int object[N][N],int target[N][N],int n)//将对象数组保存给目标数组
{
int i,j;
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
target[i][j]=object[i][j];
}
}
/* for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
printf("%d\t",tem[i][j]);
}
printf("\n");
} */
}
void update_ack(int x,int y,int n)
{
int i,j,nx,ny;
int dx[8]={-1,0,1,-1, 1,-1,0,1};//dx,dy的八对数 分别对应
int dy[8]={-1,-1,-1,0, 0, 1, 1,1 };//左上，上，右上，左，右，左下，下，右下，八个方向
attack[x][y]=1;
for(i=1;i<n;i++)
{
for(j=0;j<8;j++)
{
nx=x+i*dx[j];
ny=y+i*dy[j];
if(nx>=0 && nx<n && ny>=0 && ny<n )//nx,ny判断是否在棋盘内，满足才将对应的位标记
{
attack[nx][ny]=1;
}
}
}
// print_ack(n);测试代码
}(N can be chosen at will, if it is greater than 100, you need to modify the macro definition #define N 100 )
The result of running the code:
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