E. Road to 1600

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$rook$可以走上下左右，$queen$可以上下左右斜。每次找每个可走的最小的走。如果没走完跳跃到某个点继续走花费$1$。

构造$n\times n,rook<queen$。

  • $n<3，-1$
  • $n=3$，暴力$O(9!)$
  • $n>3$构造顺利走完除了$3\times 3$格子。
代码 
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#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define pii pair<int, int>
#define mk make_pair
const int N = 1e3 + 10;
const int mod = 1e9 + 7;
int a[N][N], v[N][N];
void dfs(int x, int y, bool qe)
{
    v[x][y] = 1;
    int minn = 1e9, ansx, ansy;
    for (int i = 1; i <= 3; i++)
        if (v[x][i] == 0)
        {
            if (minn > a[x][i])
            {
                minn = a[x][i];
                ansx = x;
                ansy = i;
            }
        }
    for (int i = 1; i <= 3; i++)
        if (v[i][y] == 0)
        {
            if (minn > a[i][y])
            {
                minn = a[i][y];
                ansx = i;
                ansy = y;
            }
        }
    if (qe)
    {
        for (int i = -3; i <= 3; i++)
        {
            int xx = x + i;
            int yy = y + i;
            if (xx >= 1 && xx <= 3 && yy >= 1 && yy <= 3 && !v[xx][yy])
            {
                if (minn > a[xx][yy])
                {
                    minn = a[xx][yy];
                    ansx = xx;
                    ansy = yy;
                }
            }
        }
        for (int i = -3; i <= 3; i++)
        {
            int xx = x + i;
            int yy = y - i;
            if (xx >= 1 && xx <= 3 && yy >= 1 && yy <= 3 && !v[xx][yy])
            {
                if (minn > a[xx][yy])
                {
                    minn = a[xx][yy];
                    ansx = xx;
                    ansy = yy;
                }
            }
        }
    }
    if (minn == 1e9)
        return;
    dfs(ansx, ansy, qe);
}
int b[10];
int vis[10];
int ok = 0;
void d(int x)
{
    if (ok)
        return;
    if (x == 10)
    {
        for (int i = 1; i <= 9; i++)
            a[(i - 1) / 3 + 1][(i - 1) % 3 + 1] = b[i];
        int ans1 = 0;
        for (int i = 1; i <= 3; i++)
            for (int j = 1; j <= 3; j++)
                v[i][j] = 0;
        for (int i = 1; i <= 3; i++)
            for (int j = 1; j <= 3; j++)
                if (v[i][j] == 0)
                    ans1++, dfs(i, j, 0);
        int ans2 = 0;
        for (int i = 1; i <= 3; i++)
            for (int j = 1; j <= 3; j++)
                v[i][j] = 0;
        for (int i = 1; i <= 3; i++)
            for (int j = 1; j <= 3; j++)
                if (v[i][j] == 0)
                    ans2++, dfs(i, j, 1);
        if (ans1 < ans2)
        {
            ok = 1;
            for (int i = 1; i <= 3; i++)
            {
                for (int j = 1; j <= 3; j++)
                    cout << a[i][j] << " ";
                cout << endl;
            }
        }

        return;
    }
    for (int i = 2; i <= 9; i++)
    {
        if (vis[i] == 0)
        {
            b[x] = i;
            vis[i] = 1;
            d(x + 1);
            vis[i] = 0;
        }
    }
}
/*
1 2 4 
5 3 8 
9 6 7  
*/
int n;
int main()
{
    // vis[1] = 1;
    // b[1] = 1;
    // d(2);
    scanf("%d", &n);
    if (n <= 2)
        return puts("-1"), 0;
    int t = n * n - 9, cur = 0;
    a[1][1] = t + 1;
    a[1][2] = t + 2;
    a[1][3] = t + 4;
    a[2][1] = t + 5;
    a[2][2] = t + 3;
    a[2][3] = t + 8;
    a[3][1] = t + 9;
    a[3][2] = t + 6;
    a[3][3] = t + 7;
    for (int i = n; i > 3; i--)
    {
        if (i & 1)
        {
            for (int j = 1; j <= i; j++)
                a[i][j] = ++cur;
            for (int j = i - 1; j >= 1; j--)
                a[j][i] = ++cur;
        }
        else
        {
            for (int j = 1; j < i; j++)
                a[j][i] = ++cur;
            for (int j = i; j >= 1; j--)
                a[i][j] = ++cur;
        }
    }
    for (int i = 1; i <= n; i++, puts(""))
        for (int j = 1; j <= n; j++)
            printf("%d ", a[i][j]);
    return 0;
}