八皇后位置放置问题

思想：主要是主对角线的元素行坐标-列坐标都相等，为保证索引为正所以加一个N，副对角线元素的行列坐标之和相等，每一行都从第一列开始遍历“`

include

using namespace std;

define N 8

int column[N+1];
int rup[2 * N + 1]; //副对角线
int lup[2 * N + 1]; //主对角线
int queen[N + 1] = { 0 };
int num = 0;

void backtrack(int);
void solution();

void main()
{
cout << “——-八皇后位置摆放——–” << endl;
int i;
for (i = 1; i <= N; i++)
column[i] = 1;
for (i = 1; i <= 2 * N; i++)
rup[i] = lup[i] = 1;
backtrack(1);
cout << “共有” << num << “种摆法” << endl;
cout << “其中一种摆法为:” << endl;
solution();

}
void backtrack(int i) //第i行的皇后
{
int j; //第i行皇后的列数
if (i > N)
num++;
else
{
for (j = 1; j <= N; j++)
{
if (column[j] == 1 && rup[i + j] == 1 && lup[i - j + N] == 1)
{
queen[i] = j;
column[j] = rup[i + j] = lup[i - j + N] = 0;
backtrack(i + 1);
column[j] = rup[i + j] = lup[i - j + N] = 1;
}
}
}
}

void solution()
{
int i, j;
for (j = 1; j <= N; j++)
{
for (i = 1; i <=N; i++)
{
if (queen[j] == i)
cout << “Q”;
else
cout << “-“;

    }
    cout << endl;
}

}
“`