CodeForces - 798B Mike and strings —— 暴力

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：给定n个字符串 每次操作可以把某个串前移一位 即最前面的字符移到最后面 问要使得所有的串都一样 最少需要移动几次

思路：枚举每一个字符串当作目标串 计算其他串移动到该串的次数 取最小即可

用vector的erase和insert 实现的题意操作

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>
#include <stack>
#include <map>
#include <cmath>
#include <vector>
#define max_ 70#define inf 0x3f3f3f3f
#define ll long long
using namespace std;
char tr[max_][max_];
vector<char>v1,v2;
int n;
int main(int argc, char const *argv[])
{
scanf("%d",&n);
int i,j,k;
for(i=1;i<=n;i++)
scanf(" %s",tr[i]);
int l=strlen(tr[1]);
int minn=inf;
for(i=1;i<=n;i++)
{
int cntsum=0;
for(j=1;j<=n;j++)
{
int cnt=0;
if(i!=j)
{
v1.clear();
v2.clear();
for(k=0;k<l;k++)
{
v1.push_back(tr[i][k]);
v2.push_back(tr[j][k]);
}
while(v1!=v2)
{
v2.insert(v2.end(),v2[0]);
v2.erase(v2.begin());
cnt++;
if(cnt>=l)
{
printf("-1\n");
return 0;
}
}
cntsum+=cnt;
}
}
if(cntsum<minn)
minn=cntsum;
}
printf("%d\n",minn );
return 0;
}