Codeforces Round #228 (Div. 2) B. Fox and Cross

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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.

这一题也是比较简单的，就是两层循环判断即可，没遇到一个‘#’就判断周围的'#'是否满足十字条件，若不满足，输出"NO"结束，否则就令这几个’#‘都为'.'，注意要对每个位置都进行判断、处理

代码如下：

#include <algorithm>
#include <iostream>
#include <string>
#include <vector>
#define maxn 110
using namespace std;char a[maxn][maxn];int main(void){int n;while(cin >> n){for(int i=0; i<n; ++i)for(int j=0; j<n; ++j)cin >> a[i][j];for(int i=0; i<n-2; ++i)for(int j=1; j<n-1; ++j){if(a[i][j] == '#'){if(a[i+1][j]=='#' && a[i+2][j]=='#' && a[i+1][j-1]=='#' && a[i+1][j+1]=='#'){a[i][j]='.';a[i+1][j]='.';a[i+2][j]='.';a[i+1][j-1]='.';a[i+1][j+1]='.';}else{cout << "NO" << endl;return 0;}}}for(int i=0; i<n; ++i)if(a[i][0]=='#' || a[i][n-1]=='#'){cout << "NO" << endl;return 0;}for(int j=1; j<n-1; ++j)if(a[n-2][j]=='#' || a[n-1][j]=='#'){cout << "NO" << endl;return 0;}cout << "YES" << endl;}return 0;
}

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