大一时老师给出的八皇后的C++代码，怎么看都看不懂：
#include <iostream> //数据输入、输出流
using std::cout; //使用名称空间std里的cout函数
using std::endl; //使用名称空间std里的endl函数
#include <iomanip> //参数化输入、输出
using std::setw; //使用名称空间std里的setw函数
#include <ctime>//定义关于时间的函数
#include <cstdlib>//C语言标准库
bool queenCheck( const char [][ 8 ], int, int );//定义queenCheck函数
void placeQueens( char [][ 8 ] );//定义placeQueens函数
void printBoard( const char [][ 8 ] );//定义printBoard函数
void xConflictSquares( char [][ 8 ], int, int );//定义xConflictSquares函数
void xDiagonals( char [][ 8 ], int, int );//定义xDiagonals函数
bool availableSquare( const char [][ 8 ] );//定义布尔类型的availableSquare函数
inline int validMove( const char board[][ 8 ], int row, int col )/*函数原型声明为内联函数（可以解决一些频繁调用的小函数大量
消耗栈空间或者是叫栈内存的问题）在行和列都在棋盘范围内的
条件下返回1，且行和列的范围是0~7*/
{ return ( row >= 0 && row < 8 && col >= 0 && col < 8 ); }//定义返回值
int main()
{
char board [ 8 ][ 8 ] = { '\0' };//初始化一个8行8列的字符型数组
srand( time( 0 ) );//给这个算法一个启动种子，也就是算法的随机种子数，有这个数以后才可以产生随机数，这个数就是当前的时间值
placeQueens( board );//调用placeQueens函数
printBoard( board ); //打印棋盘
//system("pause");//调用WINDOWS CONSOLE APP下的命令 PAUSE，暂停系统工作
return 0;
}
bool availableSquare( const char board[][ 8 ] )//遍历棋盘上所有元素，如果仍有空点，返回0，如果全放满了，返回1
{
for ( int row = 0; row < 8; ++row )
for ( int col = 0; col < 8; ++col )
if ( board[ row ][ col ] == '\0' )
return false; // at least one open square is available
return true; // no available squares
}
void placeQueens( char board[][ 8 ] )//接到board
{
const char QUEEN = 'Q';//给QUEEN赋值
int rowMove, colMove, queens = 0;//queens用来控制while循环次数
bool done = false;//初始化为0
while ( queens < 8 && !done ) {//条件：当棋盘还未满时并且queens小于8时（皇后总数不一定为8个）
//目的：随机锁定一个元素，如果条件满足，就放上皇后，并把该皇后的所在行和列以及对角线上都放上*
rowMove = rand() % 8;//得到0~8的一个随机数
colMove = rand() % 8;//得到0~8的一个随机数
if ( queenCheck( board, rowMove, colMove ) ) {//如果元素所在行、列、对角线上都没有Q
board[ rowMove ][ colMove ] = QUEEN;//就在该位置上放置Q
xConflictSquares( board, rowMove, colMove );//在锁定元素所在行、列、对角线上放置*
++queens;
}
done = availableSquare( board );//判断棋盘是否满了
}
}
void xConflictSquares( char board[][ 8 ], int row, int col )
{
for ( int loop = 0; loop < 8; ++loop ) {
// place an '*' in the row occupied by the queen，循环8次
if ( board[ row ][ loop ] == '\0' )
board[ row ][ loop ] = '*'; // place an '*' in the row（行） occupied by the queen
if ( board[ loop ][ col ] == '\0' )
board[ loop ][ col ] = '*';// place an '*' in the col（列） occupied by the queen
}
xDiagonals( board, row, col );// place an '*' in the diagonals（对角线） occupied by the queen
}
bool queenCheck( const char board[][ 8 ], int row, int col )
{
int r = row, c = col;
// check row and column for a queen
for ( int d = 0; d < 8; ++d )
if ( board[ row ][ d ] == 'Q' || board[ d ][ col ] == 'Q' )
return false;//当这行或者这列有Q，函数终止，则返回0
// check upper left diagonal for a queen
for ( int e = 0; e < 8 && validMove( board, --r, --c ); ++e )//遍历左上方对角线
if ( board[ r ][ c ] == 'Q' )
return false;//若对角线上有Q，则返回0
r = row;
c = col;
// check upper right diagonal for a queen
for ( int f = 0; f < 8 && validMove( board, --r, ++c ); ++f )//遍历右上方对角线
if ( board[ r ][ c ] == 'Q' )
return false;//若对角线上有Q，则返回0
r = row;
c = col;
// check lower left diagonal for a queen
for ( int g = 0; g < 8 && validMove( board, ++r, --c ); ++g )//遍历左下方对角线
if (board[ r ][ c ] == 'Q' )
return false;//若对角线上有Q，则返回0
r = row;
c = col;
// check lower right diagonal for a queen
for ( int h = 0; h < 8 && validMove( board, ++r, ++c ); ++h )//遍历右下方对角线
if ( board[ r ][ c ] == 'Q' )
return false;//若对角线上有Q，则返回0
return true; // no queen in conflict
}
void xDiagonals( char board[][ 8 ], int row, int col )
{
int r = row, c = col;
// upper left diagonal
for ( int a = 0; a < 8 && validMove( board, --r, --c ); ++a )//在锁定元素所在左上方对角线上放置*
board[ r ][ c ] = '*';
r = row;
c = col;
// upper right diagonal
for ( int b = 0; b < 8 && validMove( board, --r, ++c ); ++b )//在锁定元素所在右上方对角线上放置*
board[ r ][ c ] = '*';
r = row;
c = col;
// lower left diagonal
for ( int d = 0; d < 8 && validMove( board, ++r, --c ); ++d )//在锁定元素所在左下方对角线上放置*
board[ r ][ c ] = '*';
r = row;
c = col;
// lower right diagonal
for ( int e = 0; e < 8 && validMove( board, ++r, ++c ); ++e )//在锁定元素所在右上方对角线上放置*
board[ r ][ c ] = '*';
}
void printBoard( const char board[][ 8 ] )
{
int queens = 0;
// header for columns
cout << " 0 1 2 3 4 5 6 7\n";//输出列下表
for ( int r = 0; r < 8; ++r ) {
cout << setw( 2 ) << r << ' ';//输出列下表
for ( int c = 0; c < 8; ++c ) {
cout << board[ r ][ c ] << ' ';//输出改行所有元素
if ( board[ r ][ c ] == 'Q' )
++queens;
}
cout << '\n';
}
if ( queens == 8 )
cout << "\nEight Queens were placed on the board!" << endl;
else
cout << '\n' << queens << " Queens were placed on the board." << endl;
}//因为第一个皇后是随机放置的，因而不同首位置会有不同的皇后总数暑假了自己就琢磨出这么个笨办法：
#include<stdio.h>
#include<stdlib.h>
void display(int *);
int test(int *);
void main()
{
int row[8];
int count=0;//记录解决方法的个数
for(row[0]=0;row[0]<8;row[0]++)//第一行
for(row[1]=0;row[1]<8;row[1]++)//第二行
for(row[2]=0;row[2]<8;row[2]++)//第三行
for(row[3]=0;row[3]<8;row[3]++)//第四行
for(row[4]=0;row[4]<8;row[4]++)//第五行
for(row[5]=0;row[5]<8;row[5]++)//第六行
for(row[6]=0;row[6]<8;row[6]++)//第七行
for(row[7]=0;row[7]<8;row[7]++)//第八行
{
if(test(row))//测试是否满足题意，符合则返回1
{
count++;
printf("第 %-d 种放置方法:\n",count);
display(row);//显示放置方法
}
}
printf("共有 %-d 种解决方法\n",count);
}
void display(int *row)//显示放置方法
{
int i,j;
for(i=1;i<9;i++)
printf("%2d",i);//打印列表头
printf("\n");
for(i=0;i<8;i++)
{
printf("%c",i+'A');//打印行表头
for(j=0;j<row[i];j++)
printf(" #");//打印皇后前面的空白
printf(" Q");//皇后
for(j=row[i];j<7;j++)
printf(" #");//打印皇后后面的空白
printf("\n");//这一行就打印完毕，换下一行
}
}
int test(int *row)//测试是否满足题意
{
int t,i,j;
for(i=0;i<7;i++)//检查是否有皇后同列
{
t=row[i];
for(j=i+1;j<8;j++)
if(t==row[j])
return 0;
}
//到此说明没有皇后同列
for(i=0;i<7;i++)
for(j=i+1;j<8;j++)
if(row[j]==row[i]+j-i||row[j]==row[i]-(j-i))//看第i行的皇后的势力范围内是否有其他皇后
return 0;//还有其他皇后则放回0
return 1;//满足题意，返回1
}再后来学了算法设计，经过各种参考，使用回溯法终于有了这个版本：
/* 回溯法解n皇后问题(8皇后)
* file name: nqueen.c
* cmd: $gcc nqueen.c
* author: yilonglucky#gmail.com
* description: N Queens Problem with backtracking method
*/
#include<stdio.h>
#include<stdlib.h>
#include <math.h>
/* number of queens, 8 recommended */
#define QUEEN_NUM 8
#define Queen 'Q'
#define Space '#'
static int num = 0; /* solution number */
/* draw the board with queens of all solutions */
void write(int n, int *x)
{
int i, j;
printf("Solution #%d:\n", ++num);
for(i = 1; i <= n; i++)
{
/* $i is the row number */
printf("\t");
/* Queen of row i is placed in column x[i] */
for(j = 1; j < x[i]; j++)
{
/* show spaces before the queen */
printf("%c", Space);
}
/* show the queen */
printf("%c", Queen);
for(j = x[i]+1; j <= n; j++)
{
/* show spaces after the queen */
printf("%c", Space);
}
/* newline before next row*/
printf("\n");
}
/* newline before next solution */
printf("\n");
}
/* check whether a new queen
* in row k can be placed.
* return 1 if OK
* return 0 if failed
*/
int place(int k,int *x)
{
int i = 1;
/* $i means current row number,
check from line 1 to row k */
while (i < k)
{
/* compare queens in row i and row k */
if ((x[i] == x[k]) /* in the same column */
||
(abs(x[i] - x[k]) == abs(i-k))) /* in diagonal line */
{
/* queen can not be placed in column x[k] */
return 0;
}
i++; /* check the next row*/
}
/* queen in row k can be placed in column x[k] */
return 1;
}
/* N Queens Problem */
int nqueen(int n,int *x)
{
int k = 1; /* $k means current row number */
x[k] = 0; /* try to place the queen of row k in column 0 */
while (k > 0)
{
do
{
x[k]++; /* move the queen in row k to next column */
}while ((x[k] <= n) &&
(place(k,x) == 0)); /* queen in row k can be place in column x[k] */
if (x[k] <= n)
{
/* the queen in row k is still in board */
if (k == n)
{
/* if this is the last row, let's print it */
write(n, x);
}
else
{
/* ok, let's try to put a queen in next row */
x[++k] = 0;
}
}
else
{
/* backtracking: no queen can be placed in current row,
* so back to the last row.
* try to move the queen in last row to the next column. */
k--;
}
}
return 0;
}
int main()
{
int x[QUEEN_NUM+1];
int n = QUEEN_NUM;
nqueen(n, x);
getchar();
return 0;
}先贴上代码，有机会了再分析一把
