Codeforces Round #410 (Div. 2)B. Mike and strings（暴力）

传送门

Description

Mike has n strings s1, s2, ..., sn each consisting of lowercase English letters. In one move he can choose a string si, erase the first character and append it to the end of the string. For example, if he has the string "coolmike", in one move he can transform it into the string "oolmikec".

Now Mike asks himself: what is minimal number of moves that he needs to do in order to make all the strings equal?

Input

The first line contains integer n (1 ≤ n ≤ 50) — the number of strings.

This is followed by n lines which contain a string each. The i-th line corresponding to string si. Lengths of strings are equal. Lengths of each string is positive and don't exceed 50.

Output

Print the minimal number of moves Mike needs in order to make all the strings equal or print  - 1 if there is no solution.

Sample Input

4
xzzwo
zwoxz
zzwox
xzzwo

2
molzv
lzvmo

3
kc
kc
kc

3
kc
kc
kc

Sample Output

5

2

0

-1

思路

题解：

数据范围很小，因此直接暴力求解。

 

#include<bits/stdc++.h>
using namespace std;
const int INF = (1<<30)-1;
const int maxn = 55;
char str[maxn][maxn];
char tmp[maxn],tmp1[maxn];

int main()
{
	//freopen("input.txt","r",stdin);
	int n,res = INF;
	bool flag = false,success = true;
	scanf("%d",&n);
	for (int i = 0;i < n;i++)
	{
		scanf("%s",str[i]);
	}
	int len = strlen(str[0]);
	for (int i = 0;i < n;i++)
	{
		int cnt = 0;
		for (int j = 0;j < n;j++)
		{
			if (i == j)
			{
				flag = true;
				continue;
			}
			flag = false;
			if (strcmp(str[j],str[i]) == 0)
			{
				flag = true;
				continue;
			}
			strcpy(tmp1,str[j]);
			int tmplen = len;
			
			while (--tmplen)
			{
				strncpy(tmp,tmp1+1,len-1);
				tmp[len - 1] = tmp1[0];
				cnt++;
				strcpy(tmp1,tmp);
				if (strcmp(tmp1,str[i]) == 0)
				{
					flag = true;
					break;
				}
			}
			if (!flag)	
			{
				success = false;
				break;
			}
		}
		if (!flag)
		{
			success = false; 
			break;
		}
		res = min(res,cnt);
	}
	if (!success)	printf("-1
");
	else	printf("%d
",res);
	return 0;
}