Description
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.
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.
* 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'.
解题思路：首先有个结论：最小覆盖子串（串尾多一小段时，用前缀来覆盖）长度为n-prefix[n]（n-prefix[n]），n为串长。虽说此题是二维的，但是我们可以对每一行和每一列字符串都看作一个字符，将其变成一维，分别求行和列的前缀表，那么最小可覆盖子串的长度为n-prefix[n]，所以最终最小可覆盖矩阵的面积为：(row-row_prefix[row])*(col-col_prefix[col])。
AC代码：
1 #include<string.h> 2 #include<cstdio> 3 const int row_maxn=10005; 4 const int col_maxn=80; 5 int row,col,row_prefix[row_maxn],col_prefix[col_maxn]; 6 char mp[row_maxn][col_maxn],tmp[col_maxn][row_maxn]; 7 void get_row_prefix(){ 8 int i=0,j=-1; 9 row_prefix[0]=-1; 10 while(i<row){ 11 if(j==-1||!strcmp(mp[i],mp[j]))row_prefix[++i]=++j; 12 else j=row_prefix[j]; 13 } 14 } 15 void get_col_prefix(){ 16 int i=0,j=-1; 17 col_prefix[0]=-1; 18 while(i<col){ 19 if(j==-1||!strcmp(tmp[i],tmp[j]))col_prefix[++i]=++j; 20 else j=col_prefix[j]; 21 } 22 } 23 int main(){ 24 while(~scanf("%d%d",&row,&col)){ 25 for(int i=0;i<row;++i)scanf("%s",mp[i]); 26 for(int i=0;i<row;++i) 27 for(int j=0;j<col;++j) 28 tmp[j][i]=mp[i][j]; 29 get_row_prefix(); 30 get_col_prefix(); 31 printf("%d ",(row-row_prefix[row])*(col-col_prefix[col])); 32 } 33 return 0; 34 }