Milking Grid
| Time Limit: 3000MS |
|
Memory Limit: 65536K |
| Total Submissions: 3879 |
|
Accepted: 1598 |
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字符串的好题
咋一看摸不到头绪，但是自习想会有一些想法
可以求出每一行的覆盖的，和每一列的覆盖的
然后再求最小公倍数之类的处理
我们可以完善一下思路，看那个discuss里有个讲的好的
http://blog.sina.com.cn/s/blog_69c3f0410100tyjl.html
链接到这了
｛
找出每行的重复子串长度的各种可能情况，然后每行都有的并且是最小长度作为宽width。
第二步找最小重复子矩阵的高，这个思路和网上的差不多，取每行的宽为width的前缀作为一个单位，对这0到r-1个单位求出KMP的next函数，找出最小重复子序列的单位数作为高height,最终答案为width*height。
｝
代码哎
涉及了好多东西，
kmp的next 的用法 请移步这里
http://blog.csdn.net/xiaoxiaoluo/article/details/7422912这里证明了 一个字符串的最小覆盖子串是 len-next[len]
然后求那个width也有些技巧
代码很短，但很精彩
code
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <ctime>
#include <cassert>
#include <iostream>
#include <sstream>
#include <fstream>
#include <map>
#include <set>
#include <vector>
#include <queue>
#include <algorithm>
#include <iomanip>
using namespace std;
#define maxn 10005
char s[maxn][80];
int r,c;
int p[maxn],f[80];
char a[80];
int main()
{
int i,j,x,y;
scanf("%d%d",&r,&c);
memset(f,0,sizeof(f));
for(i=0; i<r; i++)
{
scanf("%s",s[i]);
strcpy(a,s[i]);
for(j=c-1; j>0; j--)
{
a[j]='\0';//changduwwei j
for(x=0,y=0; s[i][y]; x++,y++)
{
if(a[x]=='\0') x=0;//jieduan
if(a[x]!=s[i][y])break;//butong bushi chongfuzichuan
}
if(s[i][y]=='\0')f[j]++;//changduwei j keyi fugaiquanchuan
}
}
for(i=1; i<c; i++)
if(f[i]==r) break;//最短能覆盖的公共长度
x=i;
//cout<<x<<endl;
for(i=0; i<r; i++) s[i][x]='\0';
p[0]=-1;//kmp求next的过程
j=-1;
for(i=1; i<r; i++)
{
while((j!=-1)&&strcmp(s[j+1],s[i])) j=p[j];
if(strcmp(s[j+1],s[i])==0) j++;
p[i]=j;
}
//cout<<r-1-p[r-1]<<endl;
printf("%d\n",(r-1-p[r-1])*x);
return 0;
}