POJ2185 Milking Grid

POJ2185 Milking Grid

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 3220		Accepted: 1317

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'.

********************************************************************

表示题目不是很看得懂。参考了http://hi.baidu.com/%C1%D9%CA%B1%B1%B8%D3%C3%D5%CB%BA%C5/blog/item/4a047c6e417265f4421694e2.html

只不过这题让我明白了P数组可以来求一个字符串的重复字串。

#include <stdio.h>
#include <string.h>
#include <vector>
#define R 10005
#define C 80
using namespace std;

int r,c;
char gra[R][C],te[R];
int p[R],mark[R];

void KMP(int a,int b)
{
    memset(p,-1,sizeof(p));
    for(int i=1;i<b;i++)
    {
        int k=p[i-1];
        while(1)
        {
            if(te[k+1]==te[i])
            {
                p[i]=k+1;
                break;
            }
            if(k==-1)break;
            k=p[k];
        }
    }
    int k=p[b-1];
    while(1)
    {
        mark[b-1-k]++;
        if(k==-1)break;
        k=p[k];
    }
}

void re(void)
{
    scanf("%d%d",&r,&c);
    getchar();
    for(int i=0;i<r;i++)
        gets(gra[i]);
}

void run(void)
{
    memset(mark,0,sizeof(mark));
    for(int i=0;i<r;i++)
    {
        for(int j=0;j<c;j++)
            te[j]=gra[i][j];
        KMP(i,c);
    }
    int ans1,ans2;
    for(ans1=1;ans1<=c;ans1++)
        if(mark[ans1]==r)
            break;
    memset(mark,0,sizeof(mark));
    for(int i=0;i<c;i++)
    {
        for(int j=0;j<r;j++)
            te[j]=gra[j][i];
        KMP(i,r);
    }
    for(ans2=1;ans2<=r;ans2++)
        if(mark[ans2]==c)
            break;
    printf("%d\n",ans1*ans2);
}

int main()
{
    //freopen("/home/fatedayt/in","r",stdin);
    re();
    run();
    return 0;
}