CodeForces 389B Fox and Cross

链接：http://codeforces.com/problemset/problem/389/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. 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.

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.

大意——给你一个只含.和#的地图，一个#和它的上下左右的#能组成一个十字架，并且两个十字架的#不能够共用，问：图中的所有#能不能全部被十字架给覆盖掉，如果能够，输出YES，否则输出NO。注意：只有组成十字架的#才能被十字架覆盖。

思路——贪心题目。将地图全部遍历一遍即可。每一次遇到#，就判断一下它是否能被覆盖掉，如果能够，将组成十字架的地方全部改为.以免重复使用，然后继续遍历，如果不能够，则不再遍历，直接输出NO结束。这样，遍历完之后，如果全部能被覆盖就输出YES结束。检查能不能被覆盖可以这样做：假设当前#位于(x,y)，则检查(x+1,y)、(x+2,y)、(x+1,y-1)和(x+1,y+1)是不是为#，是的话就能覆盖，不是的话就不能。

复杂度分析——时间复杂度:O(n^2),空间复杂度:O(n^2)

附上AC代码：

#include <iostream>#include <cstdio>#include <string>#include <cmath>#include <iomanip>#include <ctime>#include <climits>#include <cstdlib>#include <cstring>#include <algorithm>#include <queue>#include <vector>#include <set>#include <map>//#pragma comment(linker, "/STACK:102400000, 102400000")using namespace std;typedef long long ll;const double pi = acos(-1.0);const double e = exp(1.0);const double eps = 1e-8;const int maxn = 105;char mat[maxn][maxn];int n;bool check(int x);int main(){ios::sync_with_stdio(false);while (~scanf("%d", &n)){for (int i=0; i<n; i++)scanf("%s", mat[i]);bool flag = 0;for (int i=0; i<n&&!flag; i++)for (int j=0; j<n&&!flag; j++)if (mat[i][j]=='#'){if (check(i+1) && check(i+2) && check(j-1) &&check(j+1) && mat[i+1][j-1]=='#' &&mat[i+1][j]=='#' && mat[i+1][j+1]=='#' &&mat[i+2][j]=='#'){mat[i][j] = '.';mat[i+1][j-1] = '.';mat[i+1][j] = '.';mat[i+1][j+1] = '.';mat[i+2][j] = '.';}elseflag = 1;}if (flag)printf("NO\n");elseprintf("YES\n");}return 0;}bool check(int x){if (x>=0 && x<n)return 1;return 0;}