Codeforce Round #227 Div2 B

Codeforces Round #228 (Div. 2)
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Submission	Time	Verdict
5911585	02/06/2014 08:16PM	Accepted
5911556	02/06/2014 08:11PM	Wrong answer on test 1
5911531	02/06/2014 08:06PM	Wrong answer on test 1

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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.

 1 #pragma comment(linker,"/STACK:102400000,102400000")
 2 #include <cstdio>
 3 #include <vector>
 4 #include <cmath>
 5 #include <queue>
 6 #include <set>
 7 #include <cstring>
 8 #include <iostream>
 9 #include <algorithm>
10 using namespace std;
11 #define INF 0x7fffffff 
12 #define mod 1000000007
13 #define ll long long
14 #define maxn 205
15 #define pi acos(-1.0)  
16 int n, m, k, c,t,b;
17 char s[maxn][maxn];
18 int main(){
19     scanf("%d", &n);
20     int flag = 1;
21     for (int i = 0; i < n; i++)scanf("%s", s[i]);
22     for (int i = 0; i < n; i++)
23         for (int j = 0; j < n; j++)
24             if (s[i][j] == '#'){
25                 if (j+1<n&&i+2<n&&j-1>=0&&s[i+1][j] == '#'&&s[i+2][j] == '#'&&s[i+1][j-1] == '#'&&s[i+1][j+1] == '#'){
26                     s[i+1][j] = '.';
27                     s[i+2][j] = '.';
28                     s[i+1][j-1] = '.';
29                     s[i+1][j+1] = '.';
30                     s[i][j] = '.';
31                 }
32                 else flag = 0;
33             }
34     printf(flag ? "YES\n" : "NO\n");
35     return 0;
36 }

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posted @ 2014-02-07 04:46 HaibaraAi 阅读(123) 评论(0) 收藏举报

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