Milking Grid

Milking Grid

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 9077 Accepted: 3958

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. 

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 5ABABAABABA

Sample Output

2

Hint

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

Source

USACO 2003 Fall

题意：求最小矩阵的大小

使用两次kmp 求出行的循环节  和 列的循环节 乘积即使答案

代码如下：

#include<iostream>#include<stdio.h>#include<string.h>#include<algorithm>char s[10010][80],c[80][10010];int next_s[10010],next_c[80];using namespace std;int Get_nexts(int &len){    int i=0,j=-1;    next_s[0]=-1;    while(i<len)    {        if(j==-1||strcmp(s[i],s[j])==0)            next_s[++i]=++j;        else            j=next_s[j];    }    return len-next_s[len];}int Get_nextc(int &len){    int i=0,j=-1;    next_c[0]=-1;    while(i<len)    {        if(j==-1||strcmp(c[i],c[j])==0)            next_c[++i]=++j;        else            j=next_c[j];    }    return len-next_c[len];}int main(){    int n,m;    scanf("%d%d",&n,&m);    for(int i=0;i<n;i++)        scanf("%s",&s[i]);    memset(c,'\0',sizeof(c));    for(int i=0;i<n;i++)    {       for(int j=0;j<m;j++)           c[j][i]=s[i][j];    }    printf("%d\n",Get_nexts(n)*Get_nextc(m));    return 0;}