codeforces 515D Drazil and Tiles（构造）

最新推荐文章于 2017-10-04 18:44:00 发布

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探讨了如何解决一个n*m的矩阵上使用1*2的多米诺骨牌覆盖所有黑色点的问题，并确保解的唯一性。通过算法实现，讨论了度数为1的点在寻找解决方案中的作用。

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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.

Examples

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".

题目大意：给定一个n*m的黑白矩阵，问是否存在唯一的方案使得1*2的多米诺骨牌不重叠且完全覆盖黑色点。
n,m<=1000。

题解：每次将度数为1的点取出来构造一个多米诺，直到没有度数为1的点。
若此时黑色格子已经被完全覆盖，则找到唯一解，否则有多解。
证明：
将多米诺覆盖的两个点用黑边相连，其余用白边相连，由于这是一个存在环的二分图，我们显然可以将黑边与白边交换颜色，得到另一种解。

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cstdio>
#define N 2003
using namespace std;
int n,m;
int ans[N][N],vis[N][N],size[N][N],map[N][N];
int st[2000003],st1[2000003],top;
int x[10]={0,0,-1,1},y[10]={-1,1,0,0};
void solve(int x1,int y1)
{
	for (int k=0;k<4;k++)
	{
	  	int nowx=x1+x[k]; int nowy=y1+y[k];
	  	if (nowx>0&&nowy>0&&nowx<=n&&nowy<=m&&map[nowx][nowy])
		 {
		   size[nowx][nowy]--;
		   if (size[nowx][nowy]==1&&!vis[nowx][nowy])  st[++top]=nowx,st1[top]=nowy;
	     }
	}
}
int main()
{
	scanf("%d%d",&n,&m);
	for (int i=1;i<=n;i++)
	{
		char s[N]; scanf("%s",s+1);
		for (int j=1;j<=m;j++)
		 if (s[j]=='.')  map[i][j]=1;
	}
	for (int i=1;i<=n;i++)
	 for (int j=1;j<=m;j++)
	  if (map[i][j])
	  for (int k=0;k<4;k++)
	  {
	  	int nowx=i+x[k]; int nowy=j+y[k];
	  	if (nowx>0&&nowy>0&&nowx<=n&&nowy<=m&&map[nowx][nowy])  
	  	 size[i][j]++;
	  }
	for (int i=1;i<=n;i++)
	 for (int j=1;j<=m;j++)
	  if (size[i][j]==1)  st[++top]=i,st1[top]=j;
    while (true)
    {
    	int xx=0; int yy=0;
    	while (vis[st[top]][st1[top]]&&top>0)  
    	  top--;
    	if (!top)  break;
    	xx=st[top]; yy=st1[top]; top--;
    	if (!xx&&!yy) break;
        int t=-1;
    	for (int k=0;k<4;k++)
	     {
	  	   int nowx=xx+x[k]; int nowy=yy+y[k];
	  	   if (nowx>0&&nowy>0&&nowx<=n&&nowy<=m&&map[nowx][nowy]&&!vis[nowx][nowy])
			 {
			 	t=k;
			 	break;
			 }  
	     }
	   if (t==-1)  continue;
	   int nowx=xx+x[t]; int nowy=yy+y[t];
	   vis[nowx][nowy]=1; vis[xx][yy]=1;
	   ans[nowx][nowy]=t+1; 
	   solve(nowx,nowy); solve(xx,yy);
	   if (ans[nowx][nowy]==1)  ans[xx][yy]=2;
	   if (ans[nowx][nowy]==2)  ans[xx][yy]=1;
	   if (ans[nowx][nowy]==3)  ans[xx][yy]=4;
	   if (ans[nowx][nowy]==4)  ans[xx][yy]=3; 
    }
    bool f=true;
    for (int i=1;i<=n;i++)
     for (int j=1;j<=m;j++) 
      if (map[i][j]&&!vis[i][j])  {
      	f=false;
      	break;
      }
    if (!f) {
    	printf("Not unique\n");
    	return 0;
    }
    for (int i=1;i<=n;i++)
    {
    	for (int j=1;j<=m;j++)
    	{
    		if (!map[i][j]) printf("*");
    		if (ans[i][j]==1)  printf("<");
    		if (ans[i][j]==2)  printf(">");
    		if (ans[i][j]==3)  printf("^");
    		if (ans[i][j]==4)  printf("v");
    	}
    	printf("\n");
    }
}