Codeforces Round #292 (Div. 2) D. Drazil and Tiles [拓扑排序 dfs]

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D. Drazil and Tiles

time limit per test 2 seconds

memory limit per test 256 megabytes

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 use1 × 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 solutionwhen 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 with1 × 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 test(s)

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

*

Note

In the first case, there are indeed two solutions:

<>^
^*v
v<>

and

^<>
v*^
<>v

so the answer is "Not unique".

题意及题解转自田神：http://blog.csdn.net/tc_to_top/article/details/43876015

题目大意：n*m的矩阵，' . '表示位置空，' * '表示障碍物，问能不能用尖括号填满空的点，使矩阵中所有的尖括号都两两配对，水平配对：<> 竖直配对：^v，若不存在或答案不唯一输出Not unique

题目分析：有趣的题，DFS搜索，策略：先把*的点的数量记下来，每次向四周扩展时先找只有一个方向可扩展的点扩展，因为它的灵活度最小，也就是说在它这的策略是唯一的，每个点都搜一次，最后如果n * m = cnt表示每个点都填满了（包括障碍物）则说明有且只有一解

这题还可以用拓扑排序，而不是dfs，第一份代码是我的拓扑排序，第二份是田神的dfs，结果dfs还要快，晕= =

9954592

2015-02-22 04:53:19

njczy2010

D - Drazil and Tiles

GNU C++

Accepted

233 ms

26400 KB

 #include<iostream>
 #include<cstring>
 #include<cstdlib>
 #include<cstdio>
 #include<algorithm>
 #include<cmath>
 #include<queue>
 #include<map>
 #include<set>
 #include<stack>
 #include<string>

 #define N 2010
 #define M 10005
 //#define mod 10000007
 //#define p 10000007
 #define mod2 1000000000
 #define ll long long
 #define ull unsigned long long
 #define LL long long
 #define eps 1e-6
 //#define inf 2147483647
 #define maxi(a,b) (a)>(b)? (a) : (b)
 #define mini(a,b) (a)<(b)? (a) : (b)

 using namespace std;

 int n;
 int m;
 char s[N][N];
 int r[N][N];
 int dirx[]={,,-,};          //d,r,u,l
 int diry[]={,,,-};
 int tot;

 typedef struct
 {
     int x;
     int y;
 }PP;

 int calr(int x,int y)
 {
     r[x][y]=;
     int dir;
     if(s[x][y]!='.') return -;
     int i;
     int nx,ny;
     for(i=;i<;i++){
         nx=x+dirx[i];
         ny=y+diry[i];
         if(s[nx][ny]=='.'){
             r[x][y]++;
             dir=i;
         }
     }
     return dir;
 }

 void ini()
 {
     int i,j;
     memset(r,,sizeof(r));
     for(i=;i<=n+;i++){
         for(j=;j<=m+;j++){
             s[i][j]=;
         }
     }
     for(i=;i<=n;i++){
         scanf("%s",s[i]+);
     }
     tot=;
 }

 void solve()
 {
     queue<PP>q;
     int i,j;
     PP te,nt,ntt;
     int dir;
     for(i=;i<=n;i++){
         for(j=;j<=m;j++){
             if(s[i][j]=='*'){
                 tot++;continue;
             }
             calr(i,j);
             te.x=i;te.y=j;
             if(r[i][j]==){
                 q.push(te);
             }
         }
     }
     while(q.size()>=)
     {
         te=q.front();
         q.pop();
         dir=calr(te.x,te.y);
         if(r[te.x][te.y]!=) continue;
         nt.x=te.x+dirx[ dir ];
         nt.y=te.y+diry[ dir ];
         tot+=;
         if(dir==){
             s[te.x][te.y]='^';s[nt.x][nt.y]='v';
         }
         else if(dir==){
             s[te.x][te.y]='v';s[nt.x][nt.y]='^';
         }
         else if(dir==){
             s[te.x][te.y]='<';s[nt.x][nt.y]='>';
         }
         else if(dir==){
             s[te.x][te.y]='>';s[nt.x][nt.y]='<';
         }
         for(j=;j<;j++){
             ntt.x=nt.x+dirx[j];
             ntt.y=nt.y+diry[j];
             calr(ntt.x,ntt.y);
             if(r[ntt.x][ntt.y]==){
                 q.push(ntt);
             }
         }
     }
 }

 void out()
 {
     if(tot!=n*m){
         printf("Not unique\n");
     }
     else{
         int i;
         for(i=;i<=n;i++){
             printf("%s\n",s[i]+);
         }
     }
 }

 int main()
 {
     //freopen("data.in","r",stdin);
     //freopen("data.out","w",stdout);
     //scanf("%d",&T);
     //for(int ccnt=1;ccnt<=T;ccnt++)
     //while(T--)
     //scanf("%d%d",&n,&m);
     while(scanf("%d%d",&n,&m)!=EOF)
     {
         ini();
         solve();
         out();
     }
     return ;
 }

下面转一下田神的dfs：

9954596

2015-02-22 04:54:37

njczy2010

D - Drazil and Tiles

GNU C++

Accepted

155 ms

4056 KB

 #include <cstdio>
 #include <cstring>
 int const MAX = ;
 char s[MAX][MAX];
 int n, m, cnt;
 int dx[] = {, , , -};
 int dy[] = {, -, , };

 void dfs(int x, int y)
 {
     if(x <  || x > n || y <  || y > m || s[x][y] != '.')
         return;
     int dir = -, sum = ;
     for(int i = ; i < ; i++)
     {
         int xx = x + dx[i];
         int yy = y + dy[i];
         if(s[xx][yy] == '.')
         {
             sum ++;
             dir = i;
         }
     }
     if(sum == )  //保证解的唯一性
     {
         if(dir == )
         {
             s[x][y] = '<';
             s[x][y + ] = '>';
         }
         else if(dir == )
         {
             s[x][y] = '>';
             s[x][y - ] = '<';
         }
         else if(dir == )
         {
             s[x][y] = '^';
             s[x + ][y] = 'v';
         }
         else if(dir == )
         {
             s[x][y] = 'v';
             s[x - ][y] = '^';
         }
         cnt += ;
         for(int i = ; i < ; i++)
             dfs(x + dx[dir] + dx[i], y + dy[dir] + dy[i]);
     }
 }

 int main()
 {
     cnt = ;
     scanf("%d %d", &n, &m);
     for(int i = ; i <= n; i++)
         scanf("%s", s[i] + );
     for(int i = ; i <= n; i++)
         for(int j = ; j <= m; j++)
             if(s[i][j] == '*')
                 cnt ++;
     for(int i = ; i <= n; i++)
         for(int j = ; j <= m; j++)
             dfs(i, j);
     if(cnt == n * m)
         for(int i = ; i <= n; i++)
             printf("%s\n", s[i] + );
     else
         printf("Not unique\n");
 }