Codeforces Round #297 (Div. 2) D - Arthur and Walls （深搜）

D. Arthur and Walls

time limit per test
2 seconds

memory limit per test
512 megabytes

input
standard input

output
standard output

Finally it is a day when Arthur has enough money for buying an apartment. He found a great option close to the center of the city with a nice price.

Plan of the apartment found by Arthur looks like a rectangle n × m consisting of squares of size 1 × 1.
Each of those squares contains either a wall (such square is denoted by a symbol "*" on the plan) or a free space (such square is denoted on the plan by a symbol ".").

Room in an apartment is a maximal connected area consisting of free squares. Squares are considered adjacent if they share a common side.

The old Arthur dream is to live in an apartment where all rooms are rectangles. He asks you to calculate minimum number of walls you need to remove in order to achieve this goal. After removing a wall from a square it becomes a
free square. While removing the walls it is possible that some rooms unite into a single one.

Input

The first line of the input contains two integers n, m (1 ≤ n, m ≤ 2000)
denoting the size of the Arthur apartments.

Following n lines each contain m symbols
— the plan of the apartment.

If the cell is denoted by a symbol "*" then it contains a wall.

If the cell is denoted by a symbol "." then it this cell is free from walls and also this cell is contained in some of the rooms.

Output

Output n rows each consisting of m symbols
that show how the Arthur apartment plan should look like after deleting the minimum number of walls in order to make each room (maximum connected area free from walls) be a rectangle.

If there are several possible answers, output any of them.

Examples

input

5 5
.*.*.
*****
.*.*.
*****
.*.*.

output

.*.*.
*****
.*.*.
*****
.*.*.

input

6 7
***.*.*
..*.*.*
*.*.*.*
*.*.*.*
..*...*
*******

output

***...*
..*...*
..*...*
..*...*
..*...*
*******

input

4 5
.....
.....
..***
..*..

output

.....
.....
.....
.....

以前见到过好几次这种类型的题，但是都没有思路，今天看了别
c7c3
人的代码，感觉很巧妙；找相邻的四个点，如果有一个点是×，就把它转换；

#include<stdio.h>
#include<string.h>
char mp[2100][2100];
int n,m;
void dfs(int a,int b)
{
int sum=0,i,j;
for(i=0;i<2;i++)
{
for(j=0;j<2;j++)
{
if(mp[a+i][b+j]=='*')  //找出相邻四个点中*的个数；
sum++;
}
}
if(sum==1)   //如果等于1，就“吞掉”
{
for(i=0;i<2;i++)
{
for(j=0;j<2;j++)
{
mp[a+i][b+j]='.';
}
}
for(i=-1;i<2;i++)
{
for(j=-1;j<2;j++)   //找周围8个方向继续深搜；
{
if(a+i>=0&&a+i<n-1&&b+j>=0&&b+j<m-1)
dfs(a+i,b+j);
}
}
}
}
int main()
{
int i,j,k,f;
scanf("%d%d",&n,&m);
for(i=0;i<n;i++)
{
scanf("%s",&mp[i]);
}
for(i=0;i<n-1;i++)
{
for(j=0;j<m-1;j++) //不用到边，因为后面的深搜可以覆盖到；
{
dfs(i,j);
}
}
for(i=0;i<n;i++)
{
printf("%s\n",mp[i]);
}
return 0;
}