cf228 div2 B. Fox and Cross ( 贪心， 模拟）

题目给出一个图， 问图中所有的“#”能否恰好独立的组成十字架（一个#只能在一个十字架中）一开始用ｄｆｓ写的好混乱。。　后面发现从左上到右下，如果一个＃满足正好在中间且四周可以消去，那就一定要消去，否则就ＮＯ。这样枚举一遍就好了。

题目：

B. Fox and Cross

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

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.

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.

代码：

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <vector>
#include <queue>
#include <stack>
#include <cmath>
#include <cstring>
#include <string>
#include <cstdlib>
using namespace std;
#define LL lolng long

int n;
char G[105][105];
void init()
{
    scanf("%d",&n);
    getchar();

    for(int i=0;i<n;i++)
    {
        for(int j=0;j<n;j++)
        {
            scanf("%c",&G[i][j]);
        }
        getchar();
    }
}

bool inmap(int x,int y)
{
    if(x>=0&&x<n &&y>=0&&y<n)
        return true;
    return false;
}

int dx[]= {1,-1,0,0};
int dy[] ={0,0,1,-1};


int main()
{
    init();
    int cnt=0;
    int mm=0;
    for(int i=0;i<n;i++)
    {
        for(int j=0;j<n;j++)
        {
            if( G[i][j] =='#')
            {
                mm++;
                cnt =1;
                for(int p=0;p<4;p++)
                {
                    if(inmap(i+dx[p],j+dy[p])&&G[i+dx[p]][j+dy[p]]=='#')
                        cnt++;
                }
                if(cnt==5)
                {
                    G[i][j]='.';
                    for(int p=0;p<4;p++)
                    {
                        G[i+dx[p]][j+dy[p]]='.';
                    }
                }
                //flag = dfs(i,j);
            }
        }

    }
    cnt=0;
     for(int i=0;i<n;i++)
    {
        for(int j=0;j<n;j++)
        {
                if(G[i][j] =='#')
                {
                    cnt=1;
                    break;
                }
        }
    }
    if(!cnt||mm==0)cout<<"YES"<<endl;
    else
        cout<<"NO"<<endl;
    return 0;
}