题意：
构造方格1~n*n，皇后和车的走法如图，皇后和车会走当前能走的位置中没走过的值最小的来走，如果没有花费就会加1并且跳到整个图中没走过中的最小的方格来走，让你构造一个矩阵,使得车的花费比皇后少。
思路：
n<=2显然是无解的，n=3可以通过打表打出来，然后想想怎么拓展，我们可以通过走蛇形矩阵的方法，先蛇形递增的摆放食物诱导皇后和车同轨迹走，最后让车和皇后走到n,1或者1，n，(取决于n的奇偶性），然后皇后会跳进子问题n=3的陷阱，那么皇后就输了。
代码:
// Problem: E. Road to 1600
// Contest: Codeforces - Codeforces Round #632 (Div. 2)
// URL: https://codeforces.com/contest/1333/problem/E
// Memory Limit: 256 MB
// Time Limit: 1000 ms
//
// Powered by CP Editor (https://cpeditor.org)
//#pragma GCC target("avx")
//#pragma GCC optimize(2)
//#pragma GCC optimize(3)
//#pragma GCC optimize("Ofast")
// created by myq
#include<iostream>
#include<cstdlib>
#include<string>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<climits>
#include<cmath>
#include<cctype>
#include<stack>
#include<queue>
#include<list>
#include<vector>
#include<set>
#include<map>
#include<sstream>
#include<unordered_map>
#include<unordered_set>
using namespace std;
typedef long long ll;
#define x first
#define y second
typedef pair<int,int> pii;
const int N = 510;
int a[N][N];
const int mod=998244353;
inline int read()
{
int res=0;
int f=1;
char c=getchar();
while(c>'9' ||c<'0')
{
if(c=='-') f=-1;
c=getchar();
}
while(c>='0'&&c<='9')
{
res=(res<<3)+(res<<1)+c-'0';
c=getchar();
}
return res;
}
int main()
{
int n;
cin>>n;
if(n<=2) puts("-1");
else if(n==3)
{
int eps=0;
a[1][1]=1+eps;
a[1][2]=7+eps;
a[1][3]=9+eps;
a[2][1]=3+eps;
a[2][2]=2+eps;
a[2][3]=5+eps;
a[3][1]=4+eps;
a[3][2]=8+eps;
a[3][3]=6+eps;
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
printf("%d ",a[i][j]);
cout<<endl;
}
}
else
{
int sx=4;
int sy=1;
int len=4;
int now=2;
int eps=1;
a[4][1]=1;
while(true)
{
while(sy+1<=len)
{
a[sx][++sy]=now++;
eps++;
}
while(sx-1>=1)
{
a[--sx][sy]=now++;
eps++;
}
sy++;
len++;
if(eps==n*n-9)
break;
eps++;
a[sx][sy]=now++;
while(sx+1<=len)
{
a[++sx][sy]=now++;
eps++;
}
while(sy-1>=1)
{
a[sx][--sy]=now++;
eps++;
}
if(eps==n*n-9)
break;
len++;
sx++;
eps++;
a[sx][sy]=now++;
}
a[1][1]=1+eps;
a[1][2]=7+eps;
a[1][3]=9+eps;
a[2][1]=3+eps;
a[2][2]=2+eps;
a[2][3]=5+eps;
a[3][1]=4+eps;
a[3][2]=8+eps;
a[3][3]=6+eps;
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
printf("%d ",a[i][j]);
cout<<endl;
}
}
return 0;
}
/**
* In every life we have some trouble
* When you worry you make it double
* Don't worry,be happy.
**/
