Fox and Cross

最新推荐文章于 2021-08-08 03:27:36 发布

原创最新推荐文章于 2021-08-08 03:27:36 发布 · 677 阅读

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本文介绍了一个编程挑战，任务是在一个由点和十字组成的网格上绘制特定数量的十字图案，每个十字占据五个单元格，且不能有交叉重叠。文章提供了一个C语言实现的例子，并讨论了实现过程中的关键步骤。

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. Thei-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.

思路：这个题只需看是否满足一个十字，如果满足，就将满足的十字赋值为“.”，再继续向下判断。。

就像当str[i][j]是‘#’，时，判断是否str[i+1][j-1],str[i+1][j],str[i+1][j+1],str[i+2][j]都为‘#’，如果是，赋值为‘.’！

注意：1.防止数组溢出。

2.这也是我好犯的错误，当二维数组是字符型是，记住‘\n’也属于，所以不要忘记getchar()~

#include<stdio.h>
int main()
{
    char a[200][200];
    int i,j,m,n;
    while(scanf("%d\n",&n)!=EOF)
    {
        for(i=0; i<n; i++)
        {
            for(j=0; j<n; j++)
            {
                scanf("%c",&a[i][j]);
            }
            getchar();
        }
        int ss=0;
        for(i=0; i<n; i++)
        {
            for(j=0; j<n; j++)
            {
                if(a[i][j]=='#')
                    ++ss;
            }
        }
        m=ss/5;
        if(ss==0) printf("YES\n");
        else if(ss%5!=0) printf("NO\n");
        else
        {
            int s=0;
            for(i=0; i<n-2; i++)
            {
                for(j=1; j<n; j++)
                {
                    if(a[i][j]=='#')
                    {
                        if(a[i+1][j-1]=='#'&&a[i+1][j]=='#'&&a[i+1][j+1]=='#')
                        {
                            if(a[i+2][j]=='#')
                            {
                                ++s;
                                a[i][j]='.';
                                a[i+1][j-1]='.';
                                a[i+1][j]='.';
                                a[i+1][j+1]='.';
                                a[i+2][j]='.';
                            }
                        }
                    }
                }
            }
            if(s==m)
                printf("YES\n");
            else
                printf("NO\n");
        }
    }
    return 0;
}