The eight queens problem is an old and well-known problem and a typical case of backtracking algorithms. This question was raised by the international chess player Max Bethel in 1848:
Place eight queens on an 8×8 chess board so that they cannot attack each other, that is, no two queens can be in the same row, column, or diagonal line. Ask how many ways there are.
#include
#include
#include
using namespace std;
class Queen
{
public:
int count;
Queen(int num);
~Queen();
void showQueen(int n);
void setQueen(int x,int y);
bool isClash();
bool checkClash(int nRow);
void enumQueenPostion(int &nSloution);
void putQueen(int nRow, int &nSloution);
private:
//每一个数组元素代表一个皇后的位置,queenArr[0] = 4;第一行第五格
int len;
int *queenArr;
};
Queen::Queen(int num)
{
this->len = num;
this->queenArr = (int*)malloc(sizeof(int)*num);
memset(this->queenArr,0,sizeof(int)*num);
}
Queen::~Queen()
{
delete[]queenArr;
}
//打印每一种可能情况
void Queen::showQueen(int n)
{
cout << "***************Queen position,sloution:"<= len || y < 0 || y >= len)
{
return;
}
queenArr[x] = y;
}
//判断冲突
bool Queen::isClash()
{
for (int i = 1; i <= len; ++i)
{
for (int j = 0; j <= i - 1; j++)
{
if (queenArr[i] == queenArr[j]) //说明在同一列相同值,不满足条件
{
return true;
}
if(abs(queenArr[i] - queenArr[j]) == abs(i - j)) //不可成45度角排列
{
return true;
}
}
}
return false;
}
//枚举法解决八皇后问题,枚举所有皇后可能所在的位置,将每一种可能情况列举出来
/*局限:
1.效率太低
2.不能解决n皇后问题
*/
void Queen::enumQueenPostion(int &nSloution)
{
for (queenArr[0] = 0; queenArr[0] < 8; queenArr[0]++)
for (queenArr[1] = 0; queenArr[1] < 8; queenArr[1]++)
for (queenArr[2] = 0; queenArr[2] < 8; queenArr[2]++)
for (queenArr[3] = 0; queenArr[3] < 8; queenArr[3]++)
for (queenArr[4] = 0; queenArr[4] < 8; queenArr[4]++)
for (queenArr[5] = 0; queenArr[5] < 8; queenArr[5]++)
for (queenArr[6] = 0; queenArr[6] < 8; queenArr[6]++)
for (queenArr[7] = 0; queenArr[7] < 8; queenArr[7]++)
{
count++;
if (isClash())
{
continue;
}
else
{
nSloution++;
showQueen(nSloution);
}
}
}
//检测当前当前行的状态
bool Queen::checkClash(int nRow)
{
for (int nColumn = 0; nColumn < nRow; nColumn++)
{
if (queenArr[nColumn] == queenArr[nRow])
{
return true;
}
if (abs(queenArr[nColumn] - queenArr[nRow]) == abs(nColumn - nRow))
{
return true;
}
}
return false;
}
//回溯法
void Queen::putQueen(int nRow,int &nSloution)
{
for (int i = 0; i < len; ++i)
{
queenArr[nRow] = i; //把循环放在当前循环的位置
//检查冲突
if (!checkClash(nRow))
{
if (nRow == len - 1)
{
nSloution++;
showQueen(nSloution);
}
else
{
//下一行位置
putQueen(nRow + 1,nSloution);
}
}
}
}
int main()
{
Queen queen(8);
//queen.setQueen(0,4);
//queen.setQueen(3,3);
//queen.setQueen(6,4);
int sloution = 0;
//queen.enumQueenPostion(sloution);
//queen.showQueen(sloution);
queen.putQueen(0,sloution);
return 0;
}
In addition, if you want to better improve your programming ability, learn C language and C++ programming! Overtaking in a curve, one step faster! I may be able to help you here~
Sharing (source code, actual project video, project notes, basic introductory tutorial)
Welcome partners who change careers and learn programming, use more information to learn and grow faster than thinking about it yourself!
Programming learning:
Programming learning:
