poj 2185 Milking Grid

Milking Grid

http://poj.org/problem?id=2185

Time Limit: 3000MS		Memory Limit: 65536K

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 5
ABABA
ABABA

Sample Output

2

Hint

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

Source

USACO 2003 Fall

 

题意：

在N*M字符矩阵中找出一个最小子矩阵，使其多次复制所得的矩阵包含原矩阵，输出最小矩阵面积

样例解释：多次复制 AB

解法：KMP

将矩阵每一行看做一个单位，对这n个单位哈希后做KMP，求出最短循环节len1

将矩阵每一列看做一个单位，对这m个单位哈希后做KMP，求出最短循环节len2

答案=len1*len2

#include<cstdio>
#include<iostream>
#define scale 26
using namespace std;
int n,m;
int a[10001][76];
long long horizontal[10001],vertical[77];
int fh[10011],fv[81];
void kmp()
{
	for(int i=1;i<n;i++)
	{
		int j=fh[i];
		while(j&&horizontal[j]!=horizontal[i]) j=fh[j];
		fh[i+1]= horizontal[i]==horizontal[j] ? j+1 : 0 ;
	}
	for(int i=1;i<m;i++)
	{
		int j=fv[i];
		while(j&&vertical[i]!=vertical[j]) j=fv[j];
		fv[i+1]= vertical[i]==vertical[j] ? j+1 : 0 ;
	}
}
int main()
{
	scanf("%d%d",&n,&m);
	char c;
	for(int i=1;i<=n;i++)
	 for(int j=1;j<=m;j++)
	 {
	 	cin>>c;
	 	a[i][j]=c-'A'+1;
	 	horizontal[i-1]=horizontal[i-1]*scale+a[i][j];
	 	vertical[j-1]=vertical[j-1]*scale+a[i][j];
	 }
	kmp();
	printf("%d",(n-fh[n])*(m-fv[m]));
}