POJ2185-Milking Grid（KMP，next数组的应用）

Milking Grid

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 6317		Accepted: 2648

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

题意：r*c的字符串，问用最小的面积的字符串去覆盖它。求最小的面积

思路：能够分行分列考虑，easy想到当仅仅考虑行的时候，仅仅要把每一行看成一个字符，就能够求出关于行的next数组，然后求出最短的循环串 r-next[r] ,列也是如此，所以终于答案就是 （c-P[c]）*(r-F[r]) P,F分别为各自的next数组。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <string>
#include <algorithm>
#include <queue>
using namespace std;
const int maxn = 10000+10;
const int maxm = 80;
char mat[maxn][maxm];
char revmat[maxm][maxn];
int r,c;
int P[maxn],F[maxn];
int gcd(int a,int b) {
    if(b==0) return a;
    else return gcd(b,a%b);
}
void getP() {
    P[1] = P[0] = 0;
    for(int i = 1; i < r; i++) {
        int j = P[i];
        while(j && strcmp(mat[i],mat[j])) j = P[j];
        if(strcmp(mat[i],mat[j])==0) P[i+1] = j+1;
        else P[i+1] = 0;
    }
}
void getF() {
    F[1] = F[0] = 0;
    for(int i = 1; i < c; i++) {
        int j = F[i];
        while(j && strcmp(revmat[i],revmat[j])) j = F[j];
        if(strcmp(revmat[i],revmat[j])==0) F[i+1] = j+1;
        else F[i+1] = 0;
    }
}
void getRev() {
    for(int i = 0; i < c; i++) {
        for(int j = 0; j < r; j++) {
            revmat[i][j] = mat[j][i];
        }
    }
}
void solve() {
    int L = r-P[r],R = c - F[c];
    printf("%d\n",L*R);
}
int main(){

    while(~scanf("%d%d",&r,&c)){
        for(int i = 0; i < r; i++) scanf("%s",mat[i]);
        getP();
        getRev();
        getF();
        solve();
    }
    return 0;
}