B. Forgery（暴力）

最新推荐文章于 2024-07-24 17:23:19 发布

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最新推荐文章于 2024-07-24 17:23:19 发布

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探讨如何使用特定的绘图规则，尝试复制一个医生的签名，通过分析网格中墨水点的分布，判断是否能成功伪造。

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B. Forgery

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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×m

grid, where every cell is either filled with ink or empty. Andrey's pen can fill a 3×3

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×m

grid.

Input

The first line of input contains two integers n

and m (3≤n,m≤1000

Then n

lines follow, each contains m

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).

Examples

Input

Copy

3 3
###
#.#
###

Output

Copy

YES

Input

Copy

3 3
###
###
###

Output

Copy

NO

Input

Copy

4 3
###
###
###
###

Output

Copy

YES

Input

Copy

5 7
.......
.#####.
.#.#.#.
.#####.
.......

Output

Copy

YES

Note

In the first sample Andrey can paint the border of the square with the center in (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)

and (3,2)

we have a clear paper:
```
...
...
...
...
```
use the pen with center at (2,2)

.
```
###
#.#
###
...
```
use the pen with center at (3,2)

.
```
###
###
###
###
```

In the fourth sample Andrey can paint the borders of the squares with the centers in (3,3)

and (3,5).

題意：

要伪造医生签名，先给你医生的签名nm的网格'.'表示空白',#'表示墨水，你的笔可以这么画以一个33的网格除了中心周围你都能画墨水，问你能不能伪造签名

分析：

ans统计有医生签名有多少个墨水点，只要以n*m的网格除了最外面一圈不能画之外其他都有可能画，你就判断某个点周围八个方位是否都有墨水如果都有就可以画了，并用cnt统计墨水点的个数，统计过的用‘1’标记以免重复，最后只需判断cnt是否等于sum,等于就可以否则不可以.

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <vector>
#include <set>
#include <sstream>
#include <stack>
using namespace std;

char a[1005][1005];
int b[8][2] = {{-1,-1},{-1,0},{-1,1},{0,-1},{0,1},{1,-1},{1,0},{1,1}};
int main()
{
    int n,m;
    while(cin>>n>>m){
        int ans = 0,cnt = 0;
        for(int i = 0;i < n;i++)
            for(int j = 0;j < m;j++){
                cin>>a[i][j];
                if(a[i][j] == '#')
                    ans++;
            }
        for(int i = 1;i < n-1;i++){
            for(int j = 1;j < m-1;j++){
                int co = 0;
                for(int k = 0;k < 8;k++){
                    if(a[i+b[k][0]][j+b[k][1]] == '#' || a[i+b[k][0]][j+b[k][1]] == '1')
                        co++;
                    else
                        continue;
                }
                if(co == 8){
                    for(int k = 0;k < 8;k++){
                        if(a[i+b[k][0]][j+b[k][1]] == '#')
                            cnt++;
                        a[i+b[k][0]][j+b[k][1]] = '1';
                    }
                }
            }
        }
        //cout<<"ans: "<<ans<<endl;
        //cout<<"cnt: "<<cnt<<endl;
        if(ans == cnt) cout<<"YES"<<endl;
        else
            cout<<"NO"<<endl;

    }
    return 0;
}