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 :

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]])
} 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)
for(j=; near[N][k][j]; j++)
} 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");
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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