[poj 2185] Milking Grid 解题报告（KMP+最小循环节）

题目链接：http://poj.org/problem?id=2185

题目：

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 

 

题目大意：

给出一个矩阵，求最小的可以通过复制包含原矩阵的矩阵大小

题解：

对每一行分别求最小循环节，答案矩阵的宽度就是这些最小循环节的lcm

同理列也是这样处理

#include<algorithm>
#include<cstring>
#include<iostream>
#include<cstdio>
using namespace std;

const int N=1e4+15;
int r,c,n,m;
int next[N];
char buf[N][80],tmp[N];
int kmp(int l)
{
    next[1]=0;
    for (int i=2,j=0;i<=l;i++)
    {
        while (j&&tmp[i]!=tmp[j+1]) j=next[j];
        if (tmp[i]==tmp[j+1]) j++;
        next[i]=j; 
    }
    return l-next[l];
}
int gcd(int x,int y)
{
    if (!y) return x;
    return gcd(y,x%y);
}
int lcm(int x,int y)
{
    return x/gcd(x,y)*y;
}
int main()
{
    scanf("%d%d",&r,&c);
    for (int i=1;i<=r;i++)
    {
        scanf("%s",buf[i]+1);
    }
    int n=1,m=1;
    for (int i=1;i<=r;i++)
    {
        for (int j=1;j<=c;j++) tmp[j]=buf[i][j];
        n=lcm(n,kmp(c));
        if (n>c)
        {
            n=c;
            break;        
        }
    }
    for (int i=1;i<=c;i++)
    {
        for (int j=1;j<=r;j++) tmp[j]=buf[j][i];
        m=lcm(m,kmp(r));
        if (m>r)
        {
            m=r;
            break;
        }
    }
    printf("%d\n",n*m);
    return 0;
}

 

转载于:https://www.cnblogs.com/xxzh/p/9709891.html