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/

首先先考虑如何计算一维字符串的最小覆盖子串长度：
对于某个字符串s，它的最小覆盖子串指的是长度最小的子串p，p满足通过自身的多次重复得到q，且s为q的子串。
这个可以通过KMP算法的prefix数组(next数组)得出。最小覆盖子串长度 = n - next[n-1]

先求出每一行最小覆盖子串的长度，取所有行算出来结果的的最小公倍数，得出最小覆盖矩阵的宽度。
再求出每一列的最小覆盖子串的长度，再求最小公倍数，就可以获得最小覆盖矩阵的高度了。两个相乘就是面积。

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;

}

Read full article from 2185 -- Milking Grid