2 seconds
256 megabytes
standard input
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?
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.
Print the minimal number of moves Mike needs in order to make all the strings equal or print - 1 if there is no solution.
4
xzzwo
zwoxz
zzwox
xzzwo
5
2
molzv
lzvmo
2
3
kc
kc
kc
0
3
aa
aa
ab
-1
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： 你可以将任意字符串第一个字符移到字符串尾部，可以重复此过程，问最少多少步，使得所有字符串相等
#include <iostream>
#include <cstring>
#include <string>
using namespace std; int get_step(string a, string b){
int len = a.length();
int ans = ;
for(int i = ; i < len; i++){
if(a == b)
return ans;
//string b(b, 1);
string t = "";
t += b[];
b.erase(,);
b += t;
//cout << i << " " << b << endl;
ans++;
}
return -;
} int main(){
int n;
cin >> n;
string str[];
for(int i = ; i <= n; i++){
cin >> str[i];
}
int num = ;
for(int i = ; i <= n; i++){
int cmp = num;
num = ;
for(int j = ; j <= n; j++){
if(get_step(str[i], str[j]) == -){
cout << - << endl;
return ;
}
num += get_step(str[i], str[j]);
}
num = min(num,cmp);
}
cout << num << endl;
return ;
}