题目B--Competition - 20181010

最新推荐文章于 2024-09-13 18:42:12 发布

原创最新推荐文章于 2024-09-13 18:42:12 发布 · 275 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#模拟

ACM解题专栏收录该内容

49 篇文章

订阅专栏

探讨如何使用特定的3x3印章在空白纸上复制出任意给定的签名图案，通过判断每个内部点周围是否完全被墨水包围来决定是否可以完成复制。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目

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).

Examples

Input
3 3
###
#.#
###
Output
YES
Input
3 3
###
###
###
Output
NO
Input
4 3
###
###
###
###
Output
YES
Input
5 7
.......
.#####.
.#.#.#.
.#####.
.......
Output
YES
Note

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).

题意：模拟，用形如 x x x 的印章，判断能否印出给定形状（“ . ”代表“空”）

x . x

x x x

思路：(自己的思路飙车上了天际，一开始看不懂题目以为是细菌繁殖，不对劲又觉得是找正方形，最后看明白了题，坚定的认为这就是一道扫雷类似题目，究竟是遍历井号还是空格，怎么都不对，，）

正确思路挺简单的：

遍历每一个除边缘之外的点，若其八个方向都有井，那么以这个点为中心在另一张“空白纸”上，盖上一章，（另一个二维数组对应位置标记一下），【不管该点是“ . ”还是“ # ”，判断条件只有“八个方向都有井” 】，最后对应两个“纸”，相同则可以，不同则说明原图有不规范印章，不能被印出。

代码：

#include <cstdio>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include<algorithm>
#include <set>
#include <queue>
#include <stack>
#include<vector>
#include<map>
#include<ctime>
#define ll long long
using namespace std;
int a[1100][1100];
int b[1100][1100];
int main()
{
    int n,m;
    while(cin>>n>>m){
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        for(int i=1;i<=n;++i){
            for(int j=1;j<=m;++j){
                char ch;
                cin>>ch;
                if(ch=='#')a[i][j]=1;
            }
        }
       // for(int i=0;i<=m+1;++i)a[0][i]=a[n+1][i]=1;
        //for(int i=0;i<=n+1;++i)a[i][0]=a[i][n+1]=1;
        for(int i=2;i<n;++i){
            for(int j=2;j<m;++j){
                if(a[i-1][j-1]&&a[i-1][j]&&
                   a[i-1][j+1]&&a[i][j-1]&&
                   a[i][j+1]&&a[i+1][j-1]&&
                   a[i+1][j]&&a[i+1][j+1])
                   {
                        b[i-1][j-1]=b[i-1][j]=
                        b[i-1][j+1]=b[i][j-1]=
                        b[i][j+1]=b[i+1][j-1]=
                        b[i+1][j]=b[i+1][j+1]=1;
                }
            }
        }
        bool flag=1;
        for(int i=1;i<=n;++i)
        {
            for(int j=1;j<=m;++j)
            {
                if(b[i][j]==a[i][j])continue;
                else
                {
                    flag=0;
                    break;
                }
            }
        }
        if(flag)cout<<"YES"<<endl;
        else cout<<"NO"<<endl;

    }
    return 0;
}