AC日记——Milking Grid poj 2185

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于 2017-03-09 20:30:00 发布

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原文链接：http://www.cnblogs.com/IUUUUUUUskyyy/p/6527637.html

本文介绍了一个关于寻找二维网格中最小重复单位面积的问题，通过将每一列和每一行转换为hash值，利用KMP算法找到最小的重复单元，并给出了解决方案的源代码。

Milking Grid

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 8314		Accepted: 3586

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

Hint

The entire milking grid can be constructed from repetitions of the pattern 'AB'.

Source

USACO 2003 Fall

思路：

　　每一列都给hash成一个值，然后横着kmp，找到最小循环节长度l1；

　　每一行都给hash成一个值，然后竖着kmp，找到最小循环节长度l2；

　　然后输出l1*l2；

　　劳资re十多次，才发现题目里面是多组数据；

　　然后又re十多次，看discuss才知道r和c的范围弄反了；

　　气人简直、、、

来，上代码：

#include <cstdio>
#include <cstring>
#include <iostream>

#define che 39
#define mod 100000009LL
#define mod1 100000007LL
#define mod2 13000007LL

using namespace std;

int next[101],next_[100005],r,c,ans;

long long t[101],p[100005];

char ch[100005];

int main()
{
    while(scanf("%d%d",&r,&c)==2)
    {
        memset(t,0,sizeof(t));
        memset(p,0,sizeof(p));
        memset(next,0,sizeof(next));
        memset(next_,0,sizeof(next_));
        for(int i=1;i<=r;i++)
        {
            cin>>ch;
            for(int j=0;j<c;j++)
            {
                t[j+1]=((t[j+1]*che)%mod+((ch[j]-'0')*che))%mod1;
                p[i]=((p[i]*che)%mod+((ch[j]-'0')*che))%mod2;
            }
        }
        int i=0,j=1;next[0]=-1;
        while(j<=c)
        {
            if(i==0||t[i]==t[j]) next[j++]=i++;
            else i=next[i-1]+1;
        }
        i=0,j=1,next_[0]=-1;
        while(j<=r)
        {
            if(i==0||p[i]==p[j]) next_[j++]=i++;
            else i=next_[i-1]+1;
        }
        ans=(c-next[c])*(r-next_[r]);
        cout<<ans;
        putchar('\n');
    }
    return 0;
}