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字符串的好題
咋一看摸不到頭緒，但是自習想會有一些想法
可以求出每一行的覆蓋的，和每一列的覆蓋的
然后再求最小公倍數之類的處理
我們可以完善一下思路，看那個discuss里有個講的好的
http://blog.sina.com.cn/s/blog_69c3f0410100tyjl.html
鏈接到這了
｛
找出每行的重復子串長度的各種可能情況，然后每行都有的并且是最小長度作為寬width。
第二步找最小重復子矩陣的高，這個思路和網上的差不多，取每行的寬為width的前綴作為一個單位，對這0到r-1個單位求出KMP的next函數，找出最小重復子序列的單位數作為高height,最終答案為width*height。
｝
代碼哎
涉及了好多東西，
kmp的next 的用法 請移步這里
http://blog.csdn.net/xiaoxiaoluo/article/details/7422912這里證明了 一個字符串的最小覆蓋子串是 len-next[len]
然后求那個width也有些技巧
代碼很短，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;//最短能覆蓋的公共長度
x=i;
//cout<<x<<endl;
for(i=0; i<r; i++) s[i][x]='\0';
p[0]=-1;//kmp求next的過程
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;
}