POJ 2185 Milking Grid

利用next数组求最小重复覆盖字串长度，即x = len - next[len]。先求每行最小长度再求每行最小公倍数，纵横相乘可过。

PS:以上是错误思路，仅为理解next数组。目测这篇文章是正解：http://blog.sina.com.cn/s/blog_69c3f0410100tyjl.html

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
#define N 10005
char a
[80];
int next
;
int cans, rans;
char b
;
int r, c;
void getnext(char * sub){
int i = 0;
int j = -1;
next[0] = -1;
while(sub[i]){
if(j == -1 || sub[i] == sub[j]){
j++;
i++;
if(sub[i] == sub[j])next[i] = next[j];//此处的优化不影响结果，next[len]不可能为-1；
else next[i] = j;
}
else j = next[j];
}
}
int gcd(int a, int b){
int ret = a % b;
int mut = a * b;
while(ret){
a = b;
b = ret;
ret = a % b;
}
return mut / b;
}
int main(){
while(scanf("%d%d", &r, &c) != EOF){
for(int i = 0; i < r; i++)
scanf("%s", a[i]);
for(int i = 0; i < r; i++){
getnext(a[i]);
int x = c - next[c];
if(i == 0)cans = x;
else cans = min(c, gcd(x, cans));
}
for(int i = 0; i < c; i++){
for(int j = 0; j < r; j++){
b[j] = a[j][i];
}
b[r] = 0;
getnext(b);
int x = r - next[r];
if(i == 0)rans = x;
else rans = min(r, gcd(x, rans));
}
printf("%d\n", rans * cans);
}
return 0;
}

sss