题目描述

输入样例

4

输出样例

.Q..
...Q
Q...
..Q...Q.
Q...
...Q
.Q..

参考代码

#include <iostream>using namespace std;const int N = 20;int n;
char g[N][N];
bool col[N], dg[N], udg[N];void dfs(int u)
{// u == n 搜索到最后一层if (u == n){for (int i = 0; i < n; i++) puts(g[i]);puts("");return;}// u < nfor (int i = 0; i < n; i++){if (!col[i] && !dg[i + u] && !udg[n - u + i]){g[u][i] = 'Q';col[i] = dg[u + i] = udg[n - u + i] = true;dfs(u + 1);col[i] = dg[u + i] = udg[n - u + i] = false;g[u][i] = '.';}}
}int main()
{cin >> n;for (int i = 0; i< n; i++)for (int j = 0; j < n; j++)g[i][j] = '.';dfs(0);return 0;
}

一个一个DFS

#include <iostream>using namespace std;const int N = 20;int n;
char g[N][N];
bool row[N], col[N], dg[N], udg[N];void dfs(int x, int y, int s)
{if (y == n) y = 0, x++;if (x == n){if (s == n){for (int i = 0; i < n; i++) puts(g[i]);puts("");}return;}// 不放皇后dfs(x, y + 1, s);// 放皇后if (!row[x] && !col[y] && !dg[x + y] && !udg[x - y + n]){   g[x][y] = 'Q';row[x] = col[y] = dg[x + y] = udg[x - y + n] = true;dfs(x, y + 1, s + 1);row[x] = col[y] = dg[x + y] = udg[x - y + n] = false;g[x][y] = '.';}
}int main()
{cin >> n;for (int i = 0; i< n; i++)for (int j = 0; j < n; j++)g[i][j] = '.';dfs(0, 0, 0);return 0;
}