The main idea of the topic ：B[i, j] It means how many larger numbers are around it , Can it be used B Array constructs a A Array , If you can't output “NO SOLUTION”.

analysis ： The data scale is relatively small , You can directly enumerate the values of each point .

The code is as follows ：

```#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<math.h>
using namespace std;
const int MAXN = ;
int dir[][][] = {{},{},{{},{,},{,},{,},{,}},
{{},{,},{,,},{,},{,,},{,,,},{,,},{,},{,,},{,}}};
int near[][][] = {{},{},{{},{},{},{},{,}},{{},{},{},{},{},{,},{,},{},{,},{,}}};
int B[MAXN], A[MAXN], ok, N;
bool nBigger(int k)
{
int i, cnt = , zero=;
for(i=; dir[N][k][i]; i++)
{
if(A[k] < A[dir[N][k][i]])
cnt++;
if(!A[dir[N][k][i]])
zero++;
}
if(cnt > B[k])
return false;
if(cnt+zero < B[k])
return false;
return true;
}
void DFS(int k)
{
int i, j;
if(k == N*N+)
ok = ;
if(ok)return ;
for(i=; i<MAXN; i++)
{
A[k] = i;
if(nBigger(k) == false)
continue;
for(j=; near[N][k][j]; j++)
{
if(!nBigger(near[N][k][j]))
break;
}
if(near[N][k][j] == )
DFS(k+);
if(ok)return ;
}
A[k] = ;
}
int main()
{
scanf("%d", &N);
for(int i=; i<=N; i++)
for(int j=; j<=N; j++)
scanf("%d", &B[(i-)*N+j]);
DFS();
if(!ok)
printf("NO SOLUTION\n");
else
{
for(int i=; i<=N; i++)
for(int j=; j<=N; j++)
printf("%d%c", A[(i-)*N+j], j==N?'\n':' ');
}
return ;
}```

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