【CodeFores 798B】 Mike and strings（模拟+string）

最新推荐文章于 2024-06-05 00:21:28 发布

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最新推荐文章于 2024-06-05 00:21:28 发布

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分类专栏： CF 模拟 string

本文链接：https://blog.youkuaiyun.com/fsmm_blog/article/details/71155253

CF 同时被 3 个专栏收录

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本文探讨了一个字符串同步问题，即通过移动字符串中的字符位置使多个字符串变得相同。文章介绍了一个高效的算法来解决这一问题，并通过实例展示了如何计算最小移动次数。

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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, ..., s_n each consisting of lowercase English letters. In one move he can choose a string s_i, 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 s_i. 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

input

2
molzv
lzvmo

output

input

3
kc
kc
kc

output

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".

题目大意：给n个字符串，将第i个字符串移动xi位，使得最终所有字符串相等，求最小的移动步数和，若无法相等输出-1

思路：n最大50个，枚举每一个字符串作为基础串，求得每种情况的移动步数和，取最小值。string find大发好，比赛时忘了用这个，结果没怼出来。果然还是不太熟悉

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
using namespace std;

int main()
{
    int n,ans;
    string s,t,p;
    while(~scanf("%d",&n)){
        vector<string>v;
        ans=INF;
        for (int i=0; i<n; i++){
            cin>>s;
            v.push_back(s);
        }
        for (int i=0; i<n; i++){
            int sum=0;
            t=v[i];
            for (int j=0; j<n; j++){
                if (i!=j){
                    p=v[j];
                    p+=p;
                    int k=p.find(t);   //返回下标
                    if (k==-1){
                        sum=-1;
                        break;
                    }
                    sum+=k;
                }
            }
            if (sum != -1) ans=min(ans,sum);
        }
        if (ans==INF) printf("-1\n");
        else printf("%d\n",ans);
    }
    return 0;
}