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：可以分行分列考虑，容易想到当只考虑行的时候，只要把每一行看成一个字符，就可以求出关于行的next数组，然后求出最短的循环串 r-next[r] ,列也是如此，所以最终答案就是 （c-P[c]）*(r-F[r]) P,F分别为各自的next数组。
#include
#include
#include
#include
#include
#include
#include
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; }