## Monday, April 11, 2016

### POJ 2185 -- Milking Grid

2185 -- Milking Grid
Every morning when they are milked, the Farmer John's cows form a rectangular grid that is R (1 <= R <= 10,000) rows by C (1 <= C <= 75) columns. As we all know, Farmer John is quite the expert on cow behavior, and is currently writing a book about feeding behavior in cows. He notices that if each cow is labeled with an uppercase letter indicating its breed, the two-dimensional pattern formed by his cows during milking sometimes seems to be made from smaller repeating rectangular patterns.

Help FJ find the rectangular unit of smallest area that can be repetitively tiled to make up the entire milking grid. Note that the dimensions of the small rectangular unit do not necessarily need to divide evenly the dimensions of the entire milking grid, as indicated in the sample input below.

Input
* Line 1: Two space-separated integers: R and C

* Lines 2..R+1: The grid that the cows form, with an uppercase letter denoting each cow's breed. Each of the R input lines has C characters with no space or other intervening character.
Output
* Line 1: The area of the smallest unit from which the grid is formed
Sample Input
`2 5  ABABA  ABABA  `
Sample Output
`2  `
Hint
The entire milking grid can be constructed from repetitions of the pattern 'AB'.

http://www.fgdsb.com/2015/01/13/minimum-cover-matrix/

```
int min_cover_mat(const vector<string>& map) {

if(map.empty() || map[0].empty()) return 0;

int rows = (int)map.size(), cols = (int)map[0].size();

auto lcm = [](int a, int b){

int mul = a * b;

for(int r = a % b; r ;){

a = b;

b = r;

r = a % b;

}

return mul / b;

};

auto num_cover_substr_col = [&](int r) {

int next[cols+1], i = 0, j = -1;

next[0] = -1;

while(i < cols){

if(j == -1 || map[r][i] == map[r][j]){

++i, ++j;

next[i] = j;

} else j = next[j];

}

return i - next[i];

};

auto num_cover_substr_row = [&](int c) {

int next[rows+1], i = 0, j = -1;

next[0] = -1;

while(i < rows){

if(j == -1 || map[i][c] == map[j][c]){

++i, ++j;

next[i] = j;

} else j = next[j];

}

return i - next[i];

};

int r_nums = 1, c_nums = 1;

for(int r = 0; r < rows; r ++){

c_nums = lcm(c_nums, num_cover_substr_col(r));

if(c_nums > cols){

c_nums = cols;

break;

}

}

for(int c = 0; c < c_nums; c ++){

r_nums = lcm(r_nums, num_cover_substr_row(c));

if(r_nums > rows){

r_nums = rows;

break;

}

}

return r_nums * c_nums;

}
```