CF-798B

B. Mike and strings

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

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.
Examples
input
4
xzzwo
zwoxz
zzwox
xzzwo
output
5
input
2
molzv
lzvmo
output
2
input
3
kc
kc
kc
output
0
input
3
aa
aa
ab
output
-1
Note

In the first sample testcase the optimal scenario is to perform operations in such a way as to transform all strings into "zwoxz".

题意：

将所有字符串变成相等，只允许将最左边的字符移到最右，问最少要移多少步。

若不能使所有相等，则输出-1、

分别以每一个字符串为模板，将其他的字符串移成和当前字符串相等的情况，再找出步数最少的方案。

附AC代码：

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 
 4 const int inf=1<<30;
 5 
 6 string s[60];
 7 
 8 int main(){
 9     int n,i,j,k,m;
10     cin>>n;
11     for(i=0;i<n;i++){
12         cin>>s[i];
13     }
14     int len=s[0].size();
15     int ans=inf;
16     for(i=0;i<n;i++){
17         int cnt=0;
18         for(j=0;j<n;j++){
19             for(m=0;m<len;m++){
20                 for(k=0;k<len;k++){
21                     if(s[i][k]!=s[j][(k+m)%len])
22                     break;
23                 }
24                 if(k==len)//只有len个都相等才表明移动m个字符后两字符串相等 
25                 break;
26             }
27             if(m==len){//若m==len则表明不能匹配。 
28                 cnt=inf;
29                 break;
30             }
31             cnt+=m;
32         }
33         ans=min(ans,cnt);
34     }
35     if(ans==inf)
36     cout<<-1<<endl;
37     else
38     cout<<ans<<endl;
39     return 0;
40 }