挤奶网络POJ2185

挤奶网络POJ2185

！！！补：

这个方法是错的，但是能过……
见 这个

思路

题干在这：POJ2185
按行按列kmp，各自求出最小公倍数，和长度取最小，然后相乘就好

ac代码

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<queue>
#include<cstring>
#include<vector>
#define ll long long
#define N 2000010
using namespace std;
char squ[10010][80];
char rev[80][10010];
int nex[10010];
int kmp(char* a, int n) {
	memset(nex, 0, sizeof(nex));
	int i = 0;
	for (int k = 2; k <= n; k++) {
		while (i > 0 && a[i + 1] != a[k])
			i = nex[i];
		if (a[i + 1] == a[k])
			i++;
		nex[k] = i;
	}
	return n - nex[n];
}
int LCM(int x, int y) {
	int p = x, q = y;
	while (p != 0 && q != 0) {
		if (p > q)
			p = p % q;
		else
			q = q % p;
	}
	int gcd = max(p, q);
	return x * y / gcd;
}
int main() {
	int r, c;
	cin >> r >> c;
	for (int i = 1; i <= r; i++) {
		for (int j = 1; j <= c; j++) {
			char x = getchar();
			while (x < 'A' || x > 'Z')
				x = getchar();
			squ[i][j] = x;
			rev[j][i] = x;
		}
	}
	int llcm = 1;
	for (int i = 1; i <= r; i++) {
		llcm = LCM(llcm, kmp(squ[i], c));
	}
	int hlcm = 1;
	for (int j = 1; j <= c; j++) {
		hlcm = LCM(hlcm, kmp(rev[j], r));
	}
	cout << min(llcm, c) * min(hlcm, r);
}