1 #define HAVE_STRUCT_TIMESPEC 2 #include<bits/stdc++.h> 3 #include<tuple> 4 using namespace std; 5 int x[1007][1007],y[1007][1007]; 6 queue<pair<int,int> >q,dq; 7 char ans[1007][1007]; 8 typedef struct node{ 9 int x,y; 10 char to,from; 11 }; 12 node tle[7]; 13 int main(){ 14 ios::sync_with_stdio(false); 15 cin.tie(NULL); 16 cout.tie(NULL); 17 tle[1]={0,1,'R','L'}; 18 tle[2]={1,0,'D','U'}; 19 tle[3]={0,-1,'L','R'}; 20 tle[4]={-1,0,'U','D'}; 21 int n; 22 cin>>n; 23 for(int i=1;i<=n;++i){ 24 for(int j=1;j<=n;++j){ 25 cin>>x[i][j]>>y[i][j]; 26 if(x[i][j]==-1) 27 q.push({i,j}); 28 if(x[i][j]==i&&y[i][j]==j){ 29 ans[i][j]='X'; 30 dq.push({i,j}); 31 } 32 } 33 } 34 while(!q.empty()){ 35 pair<int,int>now=q.front(); 36 q.pop(); 37 for(int i=1;i<=4;++i){ 38 int dx=now.first+tle[i].x; 39 int dy=now.second+tle[i].y; 40 if(dx>0&&dx<=n&&dy>0&&dy<=n&&x[dx][dy]==-1){//把相邻的-1全都联通,让它们在-1的联通块里无限循环 41 ans[now.first][now.second]=tle[i].to; 42 ans[dx][dy]=tle[i].from; 43 break; 44 } 45 } 46 } 47 while(!dq.empty()){//以不用动的点为起点BFS 48 pair<int,int>now=dq.front(); 49 dq.pop(); 50 for(int i=1;i<=4;++i){ 51 int dx=now.first+tle[i].x; 52 int dy=now.second+tle[i].y; 53 if(dx>0&&dx<=n&&dy>0&&dy<=n&&x[dx][dy]==x[now.first][now.second]&&y[dx][dy]==y[now.first][now.second]&&ans[dx][dy]==0){ 54 ans[dx][dy]=tle[i].from;//有相同终点的点,把它连到原本的起点 55 dq.push({dx,dy});//加入队列,这个点已经能走到它的终点,他自己也可以成为一个中间过度点 56 } 57 } 58 } 59 for(int i=1;i<=n;++i) 60 for(int j=1;j<=n;++j) 61 if(ans[i][j]==0){ 62 cout<<"INVALID"; 63 return 0; 64 } 65 cout<<"VALID "; 66 for(int i=1;i<=n;++i){ 67 for(int j=1;j<=n;++j) 68 cout<<ans[i][j]; 69 cout<<" "; 70 } 71 return 0; 72 }