Milking Grid
| Time Limit: 3000MS | Memory Limit: 65536K | |
| Total Submissions: 9905 | Accepted: 4304 |
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'.
题意很难理解，下面列出几个数据自己感觉着理解吧。
ABABA
ABABA
2
（AB）
ABCABCAB
BCABCABC
6
（ABC
BCA）
ABCDEFAB
AAAABAAA
12
（ABCDEF
AAAABA）
题解：分别求出行的最小循环长度和列的最小循环长度，然后相乘就答案。求行的最小循环长度时，把一列看成一个整体，求出next[i]，len-next[len]就是最小循环长度，同理求列。
具体代码如下：
#include<iostream>
#include<string>
#include<vector>
using namespace std;
int r,c;
vector<string> s;
bool strcmp_c(int a,int b)
{
for(int i=0;i<r;i++)
if(s[i][a]!=s[i][b])
return 0;
return 1;
}
bool strcmp_r(int a,int b)
{
for(int i=0;i<c;i++)
if(s[a][i]!=s[b][i])
return 0;
return 1;
}
int getnext_c()
{
int next[100];
int i=-1,j=0;
next[0]=-1;
while(j<c)
if(i==-1||strcmp_c(i,j))
next[++j]=++i;
else
i=next[i];
return c-next[c];
}
int getnext_r()
{
int next[10005];
int i=-1,j=0;
next[0]=-1;
while(j<r)
if(i==-1||strcmp_r(i,j))
next[++j]=++i;
else
i=next[i];
return r-next[r];
}
int main()
{
while(cin>>r>>c)
{
for(int i=0;i<r;i++)
{
string a;
cin>>a;
s.push_back(a);
}
cout<<getnext_c()*getnext_r()<<endl;
s.clear();
}
return 0;
}