试一下这组 hack ```text 10 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 ```

```cpp //13.5分 #include<bits/stdc++.h> using namespace std; const int N = 13; int g[N][N]; //输入数据 int d[N][N]; //-1为未经过，1为已经过 int p[N*N]; //存储答案 int cnt; // 表示当前答案数量 int n, k; bool flag= true;//输出答案后，用于结束递归结束程序 void dfs(int u, int x,int y) { if(u==n*n-1&&x==n&&y==n) { for(int i = 1;i<=n*n-1;i++) cout<<p[i]; flag = false; return; } int dx[8] = {-1,-1,0,1,1,1,0,-1}; int dy[8] = {0,1,1,1,0,-1,-1,-1}; for(int i = 0;i<8;i++) { int x1 = x+dx[i]; int y1 = y+dy[i]; if(x1>=1&&x1<=n&&y1>=1&&y1<=n&&d[x1][y1]==-1) { if(g[x1][y1]==g[x][y]+1||(g[x1][y1]==0&&g[x][y]==k-1)) { //表示不交叉 if(i==1&&d[x][y+1]!=-1&&d[x-1][y]!=-1) continue; if(i==3&&d[x][y+1]!=-1&&d[x+1][y]!=-1) continue; if(i==5&&d[x+1][y]!=-1&&d[x][y-1]!=-1) continue; if(i==7&&d[x-1][y]!=-1&&d[x][y-1]!=-1) continue; d[x1][y1]=1; p[++cnt] = i; dfs(u+1,x1,y1); d[x1][y1]=-1; cnt--; if(!flag) return; } } } } int main() { cin >> n>>k; for(int i = 1;i<=n;i++) for(int j = 1;j<=n;j++) cin >> g[i][j]; memset(d, -1,sizeof d); d[1][1] = 1; dfs(0,1,1); if(flag) cout << "-1"; return 0; } ```

10 2 0 1 0 1 0 1 0 1 0 1 1 0 1 0 1 0 1 0 1 0 0 1 0 1 0 1 0 1 0 1 1 0 1 0 1 0 1 0 1 0 0 1 0 1 0 1 0 1 0 1 1 0 1 0 1 0 1 0 1 0 0 1 0 1 0 1 0 1 0 1 1 0 1 0 1 0 1 0 1 0 0 1 0 1 0 1 0 1 0 1 1 0 1 0 1 0 1 0 1 0

```cpp #include <bits/stdc++.h> using namespace std; const int N = 12; int n,k; int mp[N][N]; bool vis[N][N]; int dir[8][2] = {-1,0, -1,1, 0,1, 1,1, 1,0, 1,-1, 0,-1, -1,-1}; vector<string> res; set<pair<pair<int,int>, pair<int,int>>> s; string ans; void dfs(int x,int y,int num,int cnt) { if(x == n && y == n && cnt != n*n) { return ; } if(cnt == n*n && x == n && y == n) { res.push_back(ans); return ; } for(int i = 0;i<8;i++) { int nx = x + dir[i][0]; int ny = y + dir[i][1]; if(nx<1 || ny <1 || nx>n || ny>n || vis[nx][ny] || mp[nx][ny] != (num+1)%k) continue; if(i%2) { if(s.find({{x,ny},{nx,y}}) != s.end() || s.find({{nx,y},{x,ny}}) != s.end()) { continue; } } ans += i + '0'; s.insert({{x,y},{nx,ny}}); vis[nx][ny] = true; dfs(nx,ny,mp[nx][ny],cnt+1); vis[nx][ny] = false; s.erase({{x,y},{nx,ny}}); ans.pop_back(); } } int main() { cin>>n>>k; for(int i = 1;i<=n;i++) for(int j = 1;j<=n;j++) cin>>mp[i][j]; vis[1][1] = true; dfs(1,1,mp[1][1],1); if(res.size() == 0) { cout<<-1; return 0; } sort(res.begin(), res.end()); cout<<res[0]; return 0; } ```