CodeForces 1059B

Description

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 $\mathrm{n\times mn\times m grid,\; where\; every\; cell\; is\; either\; filled\; with\; ink\; or\; empty.\; Andrey\text{'}s\; pen\; can\; fill\; a 3\times 33\times 3 square\; without\; its\; central\; cell\; if\; it\; is\; completely\; contained\; inside\; the\; grid,\; as\; shown\; below.}$

xxx
x.x
xxx

Determine whether is it possible to forge the signature on an empty $\mathrm{n\times mn\times m grid.}$

Input

The first line of input contains two integers $\mathrm{nn and mm (3\le n,m\le 10003\le n,m\le 1000).}$

Then $\mathrm{nn lines\; follow,\; each\; contains mm characters.\; Each\; of\; the\; characters\; is\; either\; \text{'}.\text{'},\; representing\; an\; empty\; cell,\; or\; \text{'}\#\text{'},\; representing\; an\; ink\; filled\; cell.}$

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 $(2,2)(2,2) and (3,2)(3,2):$

1. we have a clear paper:

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

2. use the pen with center at $(2,2)(2,2).$

   
   ###
   #.#
   ###
   ...

3. 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 $(3,3)(3,3) and (3,5)(3,5).$

不知为何，最近总是看错题意，并且非常自信的认为题意就是那样的？

这个问题需要注意一下了，最近这两场比赛全面崩盘都与看错题意有关，是自己太急躁了吗？

题中说了，每一笔都需要是一个完成的3*3的矩阵，且样例2也说明了这一点，但我愣是没看出来，导致一直WA。

很简单，暴力就完事了。

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<queue>
#include<stack>
#include<deque>
#include<iostream>
using namespace std;
typedef long long  LL;

char con[1009][1009];
char note[1009][1009];
int way[]= {1,0,-1,0,0,1,0,-1,1,1,-1,1,-1,-1,1,-1};

int main()
{
    int i,p,j,n,m;
    int flag=0;
    scanf("%d%d",&n,&m);
    for(i=1; i<=n; i++)
    {
        for(j=1; j<=m; j++)
        {
            scanf(" %c",&con[i][j]);
            if(con[i][j]=='.')
            {
                flag++;
            }
        }
    }

    while(1)
    {

//        if(n==3&&m==3&&flag==0)
//        {
//            printf("NO
");
//            break;
//        }
        for(i=2; i<n; i++)
        {
            for(j=2; j<m; j++)
            {
                for(p=0; p<=15; p+=2)
                {
                    int x=i+way[p];
                    int y=j+way[p+1];
                    if(con[x][y]=='.'&&x>=1&&x<=n&&y>=1&&y<=m)
                        break;
                }
                if(p>15)
                {
                    for(p=0; p<=15; p+=2)
                    {
                        int x=i+way[p];
                        int y=j+way[p+1];
                        if(x>=1&&x<=n&&y>=1&&y<=m)
                        {
                            note[x][y]='#';
                        }
                    }
                }
            }
        }
        int check=0;

        for(i=1; i<=n; i++)
        {
            for(j=1; j<=m; j++)
            {
                if(con[i][j]!='.'&&con[i][j]!=note[i][j])
                {
                    check=1;
                    break;
                }
            }
            if(check==1)
                break;
        }
        if(check==1)
            printf("NO
");
        else
            printf("YES
");
        break;
    }
    return 0;
}