利用一个8个元素数组:queen[i]表示第i行的皇后所在列数;
8个元素只能从0-7中选择,每个只能一次,这就保证了皇后不处在同一列或同一行,
对于不处对角:任意两行的皇后i、j,不在对角满足|i-j|!=|queen[i]-queen[j]|
由此,可解题:queen数组的全排列,对于全排列中每一个可能性,判断不在对角的条件
//8皇后问题
//全排列+逐个判断
int g_number = 0;
void Print(vector vec)
{
int len = vec.size();
for (int i = 0; i < len; ++i)
{
cout << "( " << i << "," << vec[i] << " ) ";
}
cout << endl;
}
bool Check(vector vec)
{
int len = vec.size();
if (len < 2)
return false;
for (int i = 0; i < len; ++i)
{
for (int j = i + 1; j < len;++j)
{
int row = abs(i - j);
int col = abs(vec[i] - vec[j]);
if (row == col)
return false;
}
}
return true;
}
//全排列递归
void quanpailie(vector& vec,int number)
{
int len = vec.size();
if (len == 0)
return;
if (number == len)
{
if (Check(vec))
{
++g_number;
Print(vec);
}
return;
}
else
{
for (int i = number; i < len;++i)
{
swap(vec[i], vec[number]);
quanpailie(vec, number + 1);
swap(vec[i], vec[number]);
}
}
}
void EightQueen()
{
vector queen(8, 0);
for (int i = 0; i < 8;++i)
{
queen[i] = i;
}
quanpailie(queen,0);
}
int main()
{
EightQueen();
cout << "The total number is:" << g_number << endl;
}