Milking Grid@POJ 2185

最新推荐文章于 2020-07-26 21:20:20 发布

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本文介绍了一种算法，用于找到构成整个奶牛排列网格的最小重复矩形单元的方法。该算法通过沿x和y轴分别计算重复模式来解决二维问题，并提供了完整的代码实现。

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'.

题目分析：这道题也是要利用nex数组寻找最短循环节。区别是允许最后一段可以是残缺的，即循环次数可以不是整数。比如abcd循环2.5次就是abcdabcdab，那么len-(nex[len-1]+1)就是我们要求的最短循环节的长度。

因为这道题是二维的，解决二维矩阵的关键是我们要知道在二维矩阵中循环在x和y两个方向上是独立的，可以分别计算。

也就是沿x和y方向分别做求两次nex数组，最后两循环节长度相乘就是答案。

#include <algorithm>
#include <iostream>
#include <string>
#include <vector>
#include <cstring>
#include <cstdio>
#define MAXN 10003
using namespace std;

char s[MAXN][MAXN/100];
int nex[MAXN];
int n,m;

bool equalc(int c1,int c2){
    for(int i=0;i<n;++i){
        if(s[i][c1] != s[i][c2]) return false;
    }
    return true;
}

bool equalr(int r1,int r2){
    for(int i=0;i<m;++i){
        if(s[r1][i] != s[r2][i]) return false;
    }
    return true;
}

int get_cnex(){
    nex[0] = -1;
    int k = -1;
    for(int i=1;i<m;++i){
        while(k>-1 && !equalc(k+1,i)) k = nex[k];
        if(equalc(k+1,i)) ++k;
        nex[i] = k;
    }
    return m-(nex[m-1]+1);
}

int get_rnex(){
    nex[0] = -1;
    int k = -1;
    for(int i=1;i<n;++i){
        while(k > -1 && !equalr(k+1,i)) k = nex[k];
        if(equalr(k+1,i)) ++k;
        nex[i] = k;
    }
    return n-(nex[n-1]+1);
}


int main(){
    while(~scanf("%d%d",&n,&m)){
        for(int i=0;i<n;++i) scanf("%s",s[i]);
        int ans = get_cnex(); //沿x方向求循环节长度
        ans *= get_rnex(); //沿y方向
        cout << ans << endl;
    }
    return 0;
}