Description
Drazil created a following problem about putting 1 × 2 tiles into an n × m grid: "There is a grid with some cells that are empty and some cells that are occupied. You should use 1 × 2 tiles to cover all empty cells and no two tiles should cover each other. And you should print a solution about how to do it." But Drazil doesn't like to write special checking program for this task. His friend, Varda advised him: "how about asking contestant only to print the solution when it exists and it is unique? Otherwise contestant may print 'Not unique' ". Drazil found that the constraints for this task may be much larger than for the original task! Can you solve this new problem? Note that you should print 'Not unique' either when there exists no solution or when there exists several different solutions for the original task.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 2000). The following n lines describe the grid rows. Character '.' denotes an empty cell, and the character '*' denotes a cell that is occupied.
Output
If there is no solution or the solution is not unique, you should print the string "Not unique". Otherwise you should print how to cover all empty cells with 1 × 2 tiles. Use characters "<>" to denote horizontal tiles and characters "^v" to denote vertical tiles. Refer to the sample test for the output format example.
Sample Input
Input
3 3 ... .*. ...
Output
Not unique
Input
4 4 ..** *... *.** ....
Output
<>** *^<> *v** <><>
Input
2 4 *..* ....
Output
*<>* <><>
Input
1 1 .
Output
Not unique
Input
1 1 *
Output
*
Hint
In the first case, there are indeed two solutions: <>^ ^*v v<> and ^<> v*^ <>v so the answer is "Not unique".
Source
用贪心法解决：首先应该着力填充周围只有一个空格的点，随后以它的空格为中心，再不断向外拓展，直到整个界面被填充。
注意在judge函数判断边界和‘.’时应该这样写：return (i>=0 && i<n && j>=0 && j<m && mp[i][j]=='.'); 分开判断一直错。
1 #pragma comment(linker, "/STACK:1024000000,1024000000") 2 #include<iostream> 3 #include<cstdio> 4 #include<cstring> 5 #include<cmath> 6 #include<math.h> 7 #include<algorithm> 8 #include<queue> 9 #include<set> 10 #include<bitset> 11 #include<map> 12 #include<vector> 13 #include<stdlib.h> 14 #include <stack> 15 using namespace std; 16 int dirx[]={0,0,-1,1}; 17 int diry[]={-1,1,0,0}; 18 #define PI acos(-1.0) 19 #define max(a,b) (a) > (b) ? (a) : (b) 20 #define min(a,b) (a) < (b) ? (a) : (b) 21 #define ll long long 22 #define eps 1e-10 23 #define MOD 1000000007 24 #define N 2006 25 #define inf 1e12 26 int n,m; 27 char mp[N][N]; 28 struct Node{ 29 int x,y; 30 }; 31 char change[]="><<>v^^v"; 32 bool judge(int i,int j){ 33 return (i>=0 && i<n && j>=0 && j<m && mp[i][j]=='.'); 34 } 35 36 int nearPoint_num(int x,int y){ 37 int ans=0;//ans表示周围的空点 38 for(int i=0;i<4;i++){ 39 int tx=x+dirx[i]; 40 int ty=y+diry[i]; 41 if(judge(tx,ty)){ 42 ans++; 43 } 44 } 45 return ans; 46 } 47 48 49 void bfs(){ 50 queue<Node>q; 51 Node tmp; 52 Node t1,t2,t3; 53 for(int i=0;i<n;i++){ 54 for(int j=0;j<m;j++){ 55 if(judge(i,j) && nearPoint_num(i,j)==1){ 56 tmp.x=i; 57 tmp.y=j; 58 //printf("***%d %d ",tmp.x,tmp.y); 59 q.push(tmp); 60 } 61 } 62 } 63 while(!q.empty()){ 64 t1=q.front(); 65 q.pop(); 66 for(int i=0;i<4;i++){ 67 t2.x=t1.x+dirx[i]; 68 t2.y=t1.y+diry[i]; 69 if(judge(t2.x,t2.y)){ 70 mp[t1.x][t1.y]=change[i*2]; 71 mp[t2.x][t2.y]=change[i*2+1]; 72 //printf("%d %d %c ",t1.x,t1.y,mp[t1.x][t1.y]); 73 //printf("%d %d %c ",t2.x,t2.y,mp[t2.x][t2.y]); 74 75 for(int j=0;j<4;j++){ 76 t3.x=t2.x+dirx[j]; 77 t3.y=t2.y+diry[j]; 78 if(judge(t3.x,t3.y) && nearPoint_num(t3.x,t3.y)==1){ 79 q.push(t3); 80 } 81 } 82 83 } 84 85 } 86 } 87 88 int flag=1; 89 for(int i=0;i<n;i++){ 90 for(int j=0;j<m;j++){ 91 if(mp[i][j]=='.'){ 92 flag=0; 93 break; 94 } 95 } 96 if(flag==0){ 97 break; 98 } 99 } 100 if(flag==0){ 101 printf("Not unique "); 102 } 103 else{ 104 for(int i=0;i<n;i++){ 105 for(int j=0;j<m;j++){ 106 printf("%c",mp[i][j]); 107 } 108 printf(" "); 109 } 110 } 111 112 } 113 int main() 114 { 115 while(scanf("%d%d",&n,&m)==2){ 116 for(int i=0;i<n;i++){ 117 scanf("%s",mp[i]); 118 } 119 120 bfs(); 121 122 } 123 return 0; 124 }