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.

* 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.

* Line 1: The area of the smallest unit from which the grid is formed 

The entire milking grid can be constructed from repetitions of the pattern 'AB'.

做两次KMP

行和列分别是len-next[len];

最后两个结果相乘就可以了

 

本题只是把线性变成平面，kmp对字符的操作变成字符串……

 

 #include<iostream>

 #include<cstdio>

 #include<cstring>

 using namespace std;

 #define maxn 1000008

 char s[maxn][];

 int r, c, next[maxn];

 bool same1(int i, int j)   // 判断第i行和第j行是否相等

 {

     for(int k = ; k < c; k++)

         if(s[i][k] != s[j][k])

             return false;

     return true;

 }

 bool same2(int i, int j)  // 判断第i列和第j列是否相等。

 {

     for(int k = ; k < r; k++)

         if(s[k][i] != s[k][j])

             return false;

     return true;

 }

 int main()

 {

     while(~scanf("%d%d", &r, &c))

     {

         for(int i = ; i < r; i++)

             scanf("%s", s[i]);

         int j, k;

         memset(next, , sizeof(next));

         j = ;

         k = next[] = -;

         while(j < r)

         {

             while(- != k && !same1(j, k))

                 k = next[k];

             next[++j] = ++k;

         }

         int ans1 = r - next[r];     //  r-next[r]就是需要的最短的长度可以覆盖这个平面

         memset(next, , sizeof(next));

         j = ;

         k = next[] = -;

         while(j < c)

         {

             while(- != k && !same2(j, k))

                 k = next[k];

             next[++j] = ++k;

         }

         int ans2 = c - next[c];  //列的 

         printf("%d\n", ans1*ans2);

     }

     return ;

 }