题意大致为把
xxx x.x xxx
看作一个整体，刚开始字符全为'.',看是否可以组成给出的图像。
做法事用一个数组输入，另一个数组刚开始全赋为.........................................,当三行三列一周全为#时，把另一个数组相应位置上的点改为#，最后比较两个数组是否一模一样
Student Andrey has been skipping physical education lessons for the whole term, and now he must somehow get a passing grade on this subject. Obviously, it is impossible to do this by legal means, but Andrey doesn't give up. Having obtained an empty certificate from a local hospital, he is going to use his knowledge of local doctor's handwriting to make a counterfeit certificate of illness. However, after writing most of the certificate, Andrey suddenly discovered that doctor's signature is impossible to forge. Or is it?
For simplicity, the signature is represented as an n×mn×m grid, where every cell is either filled with ink or empty. Andrey's pen can fill a 3×33×3 square without its central cell if it is completely contained inside the grid, as shown below.
xxx x.x xxx
Determine whether is it possible to forge the signature on an empty n×mn×m grid.
Input
The first line of input contains two integers nn and mm (3≤n,m≤10003≤n,m≤1000).
Then nn lines follow, each contains mm characters. Each of the characters is either '.', representing an empty cell, or '#', representing an ink filled cell.
Output
If Andrey can forge the signature, output "YES". Otherwise output "NO".
You can print each letter in any case (upper or lower).
Sample Input
Input
3 3 ### #.# ###
Output
YES
Input
3 3 ### ### ###
Output
NO
Input
4 3 ### ### ### ###
Output
YES
Input
5 7 ....... .#####. .#.#.#. .#####. .......
Output
YES
Hint
In the first sample Andrey can paint the border of the square with the center in (2,2)(2,2).
In the second sample the signature is impossible to forge.
In the third sample Andrey can paint the borders of the squares with the centers in (2,2)(2,2) and (3,2)(3,2):
- we have a clear paper:
... ... ... ...
- use the pen with center at (2,2)(2,2).
### #.# ### ...
- use the pen with center at (3,2)(3,2).
### ### ### ###
In the fourth sample Andrey can paint the borders of the squares with the centers in (3,3)(3,3) and (3,5)(3,5).
代码：
#include <iostream>
#include<stdio.h>
#include<algorithm>
#include<stdlib.h>
#include<string.h>
#include<map>
#include<stack>
#include<vector>
#include<queue>
#include<set>
using namespace std;
#define maxn 100010
int main()
{
char a[1010][1010],b[1010][1010];
int s,m,n,k,i,j,flag;
while(scanf("%d%d",&m,&n)!=EOF)
{
for(i=0;i<m;i++)
for(j=0;j<n;j++)
{
cin>>a[i][j];
b[i][j]='.';
}
for(i=0;i<m-2;i++)
{
for(j=0;j<n-2;j++)
{
if(a[i][j]=='#'&&a[i][j+1]=='#'&&a[i][j+2]=='#'&&a[i+1][j]=='#'
&&a[i+1][j+2]=='#'&&a[i+2][j]=='#'&&a[i+2][j+1]=='#'&&a[i+2][j+2]=='#')
b[i][j]='#',b[i][j+1]='#',b[i][j+2]='#',b[i+1][j]='#'
,b[i+1][j+2]='#',b[i+2][j]='#',b[i+2][j+1]='#',b[i+2][j+2]='#';
}
}
flag=0;
for(i=0;i<m;i++)
for(j=0;j<n;j++)
if(a[i][j]!=b[i][j])
{
flag=1;break;
}
if(flag)
printf("NO\n");
else
printf("YES\n");
}
return 0;
}