Codeforces Round #472 B. Mystical Mosaic

B. Mystical Mosaic
There is a rectangular grid of n rows of m initially-white cells each.

Arkady performed a certain number (possibly zero) of operations on it. In the i-th operation, a non-empty subset of rows Ri and a non-empty subset of columns Ci are chosen. For each row r in Ri and each column c in Ci, the intersection of row r and column c is coloured black.

There’s another constraint: a row or a column can only be chosen at most once among all operations. In other words, it means that no pair of (i, j) (i < j) exists such that or , where denotes intersection of sets, and denotes the empty set.

You are to determine whether a valid sequence of operations exists that produces a given final grid.

Input
The first line contains two space-separated integers n and m (1 ≤ n, m ≤ 50) — the number of rows and columns of the grid, respectively.

Each of the following n lines contains a string of m characters, each being either ‘.’ (denoting a white cell) or ‘#’ (denoting a black cell), representing the desired setup.

Output
If the given grid can be achieved by any valid sequence of operations, output “Yes”; otherwise output “No” (both without quotes).

You can print each character in any case (upper or lower).

Examples
inputCopy
5 8
.#.#..#.
…..#..
.#.#..#.

.#….

…..#..
output
Yes
inputCopy
5 5
..#..
..#..

#

..#..
..#..
output
No
inputCopy
5 9
……..#

……..

..##.#…
…….#.
….#.#.#
output
No
Note
For the first example, the desired setup can be produced by 3 operations, as is shown below.

For the second example, the desired setup cannot be produced, since in order to colour the center row, the third row and all columns must be selected in one operation, but after that no column can be selected again, hence it won’t be possible to colour the other cells in the center column.

题意：初始矩阵全是’.’,每一行与每一列可以做交集操作，交集操作后，将某行某列相交的网格设置为’#’,每行每列只能做一次操作，求有限次操作后能否得到给定的矩阵。

思路：暴力检验同列的’#’,所在的行是否相等

代码：

#include
#include
#include
using namespace std;
typedef long long ll;
string str[105];
    int n,m;
bool ck(int i,int j){
    for(int x=0;xif(str[x][j]=='#'){
            if(str[x]!=str[i])
            return 0;
        }
    }
    return 1;
}
int main(){
cin>>n>>m;
    for(int i=0;icin>>str[i];
    }
    int len=str[0].length();

    for(int i=0;ifor(int j=0;jif(str[i][j]=='#'){
                if(!ck(i,j)){
                    cout<<"No"<return 0;
                }
            }
        }

    }

    cout<<"Yes"<return 0;
}