POJ2185 Milking Grid(KMP)

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'.

 

题意:在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

 

分析:为了找到最小重复的子矩阵，我们需要找到最小重复的宽度和长度.  

最小重复的宽度需要满足每一行，所以我们枚举了每种宽度满足的行数.

最后找到最小重复的宽度.

找最小重复的长度的时候，进行在竖直方向上，对满足最小重复宽度的前缀进行KMP的next操作

找到最小重复的长度

 

贴一下一位讲解详细的大佬：http://blog.sina.com.cn/s/blog_69c3f0410100tyjl.html

 

代码如下

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
char str[10010][80];
char a[80];
int f[10010];//用来统计最小重复长度是下标长度的行数有多少.
int Next[10010];
int r,c,i,j,p,q,kuan;
void getNext()
{
    int j, k;
    j = 0; k = -1; Next[0] = -1;
    while(j < r)
        if(k == -1 ||!strcmp(str[j],str[k])) //进行多个字符数组之间的KMP操作
            Next[++j] = ++k;
        else
            k = Next[k];

}
int main()
{
    while(scanf("%d%d",&r,&c)!=EOF)
    {
        for(int i=0;i<r;i++)
            scanf("%s",str[i]);
            memset(f,0,sizeof(f));
        for(int i=0;i<r;i++){
           memcpy(a,str[i],sizeof(str[i]));
          for(int j=c;j>0;j--)
          {
            a[j]='\0';
            for(p=0,q=0;str[i][q];p++,q++)
            {
                if(p==j)p=0;
                if(a[p]!=str[i][q])break;
            }
            if(!str[i][q])f[j]++;//能够匹配到字符串末端 即循环节成立，进行计数
          }
        }
       for(i=1;i<=c;i++)  
        {
          if(f[i]==r)  //求得满足所有行数的最小重复长度
          {
              kuan=i;
              break;
          }
        }
       for(i=0;i<r;i++)str[i][kuan]='\0';// 满足最小重复的宽度的前缀中进行KMP操作
       getNext();
       printf("%d\n",(r-Next[r])*kuan);//长*宽 即为最小的重复子矩阵
    }
    return 0;
}