問題 B: Grid Compression

時間限制: 1 Sec  記憶體限制: 128 MB

題目描述

There is a grid of squares with H horizontal rows and W vertical columns. The square at the i-th row from the top and the j-th column from the left is represented as (i,j). Each square is black or white. The color of the square is given as an H-by-W matrix (ai,j). If ai,j is ‘.‘, the square (i,j) is white; if ai,j is '#', the square (i,j) is black.
Snuke is compressing this grid. He will do so by repeatedly performing the following operation while there is a row or column that consists only of white squares:
Operation: choose any one row or column that consists only of white squares, remove it and delete the space between the rows or columns.
It can be shown that the final state of the grid is uniquely determined regardless of what row or column is chosen in each operation. Find the final state of the grid.
Constraints
1≤H,W≤100
ai,j is '.' or '#'.
There is at least one black square in the whole grid.

輸入

Input is given from Standard Input in the following format:

H W
a1,1…a1,W
:
aH,1…aH,W

輸出

Print the final state of the grid in the same format as input (without the numbers of rows and columns); see the samples for clarity.

樣例輸入

複製樣例資料

4 4
##.#
....
##.#
.#.#

樣例輸出

###
###
.##

提示

The second row and the third column in the original grid will be removed.

/**/
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cctype>
#include <iostream>
#include <algorithm>
#include <map>
#include <set>
#include <vector>
#include <string>
#include <stack>
#include <queue>

typedef long long LL;
using namespace std;

int n, m;
char s[105][105];
int l[105], r[105];

int main()
{
	//freopen("in.txt", "r", stdin);
	//freopen("out.txt", "w", stdout);

	scanf("%d %d", &n, &m);
	for (int i = 1; i <= n; i++){
		scanf("%s", s[i] + 1);
	}
	for (int i = 1; i <= n; i++){
		int t = 0;
		for (int j = 1; j <= m; j++){
			if(s[i][j] == '.'){
				t++;
			}
		}
		if(t == m) l[i] = 1;
	}
	for (int i = 1; i <= m; i++){
		int t = 0;
		for (int j = 1; j <= n; j++){
			if(s[j][i] == '.'){
				t++;
			}
		}
		if(t == n) r[i] = 1;
	}
	for (int i = 1; i <= n; i++){
		int flag = 0;
		for (int j = 1; j <= m; j++){
			if(l[i] || r[j]) continue;
			flag = 1;
			printf("%c", s[i][j]);
		}
		if(flag) printf("\n");
	}

	return 0;
}
/**/