POJ 2185 Milking Grid [KMP]

Milking Grid

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 8226		Accepted: 3549

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

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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。N<=10000,M<=75

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只需要一行做一个字母求一次，再一列做一个字母求一次就好了，然后和子串是一样的n-fail[n]

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;
typedef long long ll;
const int N=1e4+,M=;
int n,m,k,fail[N];
char s[N][M];
bool cmp1(int a,int b){
    for(int i=;i<=m;i++) if(s[a][i]!=s[b][i]) return false;
    return true;
}
bool cmp2(int a,int b){
    for(int i=;i<=k;i++) if(s[i][a]!=s[i][b]) return false;
    return true;
}
void getFail(int n,bool (*cmp)(int a,int b)){
    fail[]=;
    for(int i=;i<=n;i++){
        int j=fail[i-];
        while(j&&!cmp(j+,i)) j=fail[j];
        fail[i]=cmp(j+,i)?j+:;
    }
}

int main(){
//    freopen("in.txt","r",stdin);
    scanf("%d%d",&n,&m);
    for(int i=;i<=n;i++) scanf("%s",s[i]+);
    getFail(n,cmp1);
    k=n-fail[n];
    memset(fail,,sizeof(fail));
    getFail(m,cmp2);
    printf("%d",k*(m-fail[m]));
}