_hello_word_ @ 2024-07-10 19:09:18

#include <bits/stdc++.h>
using namespace std;
int n, cnt = 0, a[15][15] = { 0 }, ans[15];
void seta(int i, int j, int p)
{
    ans[i] = j;
    a[i][j] = 1;
    for (int k = 1; k <= n; k++) // 同行同列
    {
        a[k][j] = p;
        a[i][k] = p;
    }
    for (int m = 1; m <= n; m++)
    { // 副对角线方向
        for (int l = 1; l <= n; l++)
        {
            if (m + l == i + j)
                a[m][l] = p;
            if (m - l == i - j) // 主对角线方向
                a[m][l] = p;
        }
    }

}
void dfs(int pos)
{
    if (pos == n + 1)
    {
        cnt += 1;
        if (cnt <= 3) {
            for (int i = 1; i <= n; i++)
                cout << ans[i] << " ";
            cout << endl;
        }
    }
    else if (pos < n + 1)
    {
        for (int j = 1; j <= n; j++)
        {
            if (!a[pos][j])
            {
                int b[15][15] = { 0 };
                memcpy(b, a, sizeof(a));
                seta(pos, j, 1);
                dfs(pos + 1);
                memcpy(a, b, sizeof(a));
            }
        }
    }
}
int main()
{
    cin >> n;
    dfs(1);
    cout << cnt;
    return 0;
}

by NEKO_Daze @ 2024-07-10 19:13:08

@_helloword 你这 dfs 是不是没回溯啊？

by _hello_word_ @ 2024-07-11 08:32:18

@NEKO_Daze memcpy(a,b,sizeof(a))算不算回溯啊？虽然占空间有点大

by hzy99999 @ 2024-07-11 09:13:50

1、一次memcpy的时间复杂度是O(n^2)，这样肯定超时，不需要b数组，你直接对a数组进行操作以及回溯就行了
2、你给对角线做标记的操作是O(n^2),这样也会超时
3、打标记的时候不能只有0和1，应该标记的时候+1，回溯的时候-1，这样可以表示这个格子被多少个皇后覆盖
4、你是谁呀，我为啥关注你？ @_helloword
附个代码：

#include <bits/stdc++.h>
using namespace std;
int n, cnt = 0, a[15][15] = {0}, ans[15];
void seta(int i, int j, int p)
{
    ans[i] = j;
    a[i][j] += p;
    for (int k = 1; k <= n; k++) // 同行同列
    {
        a[k][j] += p;
        a[i][k] += p;
    }
    // for (int m = 1; m <= n; m++)
    // { // 副对角线方向
    //     for (int l = 1; l <= n; l++)
    //     {
    //         if (m + l == i + j)
    //             a[m][l] += p;
    //         if (m - l == i - j) // 主对角线方向
    //             a[m][l] += p;
    //     }
    // }
    int x = i + 1, y = j + 1;
    while (x <= n && y <= n)
        a[x][y] += p, x++, y++;
    x = i + 1, y = j - 1;
    while (x <= n && y)
        a[x][y] += p, x++, y--;
}
void dfs(int pos)
{
    if (pos == n + 1)
    {
        cnt += 1;
        if (cnt <= 3)
        {
            for (int i = 1; i <= n; i++)
                cout << ans[i] << " ";
            cout << endl;
        }
    }
    else if (pos < n + 1)
    {
        for (int j = 1; j <= n; j++)
        {
            if (!a[pos][j])
            {
                seta(pos, j, 1);
                dfs(pos + 1);
                seta(pos, j, -1);
            }
        }
    }
}
int main()
{
    cin >> n;
    dfs(1);
    cout << cnt;
    return 0;
}

by GEZHENHAO @ 2024-07-12 16:41:10

@_helloword 你把数组开大一点```

include <bits/stdc++.h>

using namespace std; int n, cnt = 0, a[30][30] = { 0 }, ans[15];//这里