Codeforces Round #228 (Div. 2)B. Fox and Cross_you've got two rectangular tables with sizes na-优快云博客

本文介绍了一种通过判断能否使用特定的十字形图案拼接覆盖一个由'#'和'.'组成的n×n矩阵的问题。该问题要求每个十字形图案覆盖五个'#'字符，并且不能有重叠。文章提供了一个C++代码示例来解决这个问题。

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Fox Ciel has a board with n rows and n columns. So, the board consists of n × n cells. Each cell contains either a symbol '.', or a symbol '#'.

A cross on the board is a connected set of exactly five cells of the board that looks like a cross. The picture below shows how it looks.

$\text{[math]}$

Ciel wants to draw several (may be zero) crosses on the board. Each cross must cover exactly five cells with symbols '#', and any cell with symbol '#' must belong to some cross. No two crosses can share a cell.

Please, tell Ciel if she can draw the crosses in the described way.

Input

The first line contains an integer n (3 ≤ n ≤ 100) — the size of the board.

Each of the next n lines describes one row of the board. The i-th line describes the i-th row of the board and consists of n characters. Each character is either a symbol '.', or a symbol '#'.

Output

Output a single line with "YES" if Ciel can draw the crosses in the described way. Otherwise output a single line with "NO".

Sample test(s)

input
5
.#...
####.
.####
...#.
.....

output
YES

input
4
####
####
####
####

output
NO

input
6
.#....
####..
.####.
.#.##.
######
.#..#.

output
YES

input
6
.#..#.
######
.####.
.####.
######
.#..#.

output
NO

input
3
...
...
...

output
YES

Note

In example 1, you can draw two crosses. The picture below shows what they look like.

$\text{[math]}$

In example 2, the board contains 16 cells with '#', but each cross contains 5. Since 16 is not a multiple of 5, so it's impossible to cover all.

解题报告：

此题解法就是是否能用若干个十字形的图案拼成。代码如下：

#include <iostream>
#include<cstdio>
#include<cstring>
using namespace std;
char s[111][111];
int vis[111][111];
int main(){
    int t,flag;
    flag=0;
    memset(vis,0,sizeof(vis));
    scanf("%d",&t);
    for(int i=0;i<t;i++)
        scanf("%s",s[i]);
    for(int i=0;i<t;i++)
        for(int j=0;j<t;j++)
            if(s[i][j]=='#'&&!vis[i][j]){
                if((s[i-1][j]=='#'&&s[i+1][j]=='#'&&s[i][j-1]=='#'&&s[i][j+1]=='#')&&(!vis[i-1][j]&&!vis[i+1][j]&&!vis[i][j-1]&&!vis[i][j+1])){
                        vis[i-1][j]=1;vis[i+1][j]=1;vis[i][j-1]=1;vis[i][j+1]=1;vis[i][j]=1;
                        s[i-1][j]='.';s[i+1][j]='.';s[i][j-1]='.';s[i][j+1]='.';s[i][j]='.';
                   }
            }
    for(int i=0;i<t;i++)
        for(int j=0;j<t;j++)
            if(s[i][j]=='#')
                flag=1;
    if(flag)
        printf("NO\n");
    else
        printf("YES\n");
    return 0;
}