时间限制:1000ms

单点时限:1000ms

内存限制:256MB

描述

Lara Croft, the fiercely independent daughter of a missing adventurer, must push herself beyond her limits when she discovers the island where her father disappeared. In this mysterious island, Lara finds a tomb with a very heavy door. To open the door, Lara must input the password at the stone keyboard on the door. But what is the password? After reading the research notes written in her father's notebook, Lara finds out that the key is on the statue beside the door.

The statue is wearing many arm rings on which some letters are carved. So there is a string on each ring. Because the letters are carved on a circle and the spaces between any adjacent letters are all equal, any letter can be the starting letter of the string. The longest common subsequence (let's call it "LCS") of the strings on all rings is the password. A subsequence is a sequence that can be derived from another sequence by deleting some or no elements without changing the order of the remaining elements.

For example, there are two strings on two arm rings: s1 = "abcdefg" and s2 = "zaxcdkgb". Then "acdg" is a LCS if you consider 'a' as the starting letter of s1, and consider 'z' or 'a' as the starting letter of s2. But if you consider 'd' as the starting letter of s1 and s2, you can get "dgac" as a LCS. If there are more than one LCS, the password is the one which is the smallest in lexicographical order.

Please find the password for Lara.

输入

There are no more than 10 test cases.

In each case:

The first line is an integer n, meaning there are n (0 < n ≤ 10) arm rings.

Then n lines follow. Each line is a string on an arm ring consisting of only lowercase letters. The length of the string is no more than 8.

输出

For each case, print the password. If there is no LCS, print 0 instead.

样例输入

2

abcdefg

zaxcdkgb

5

abcdef

kedajceu

adbac

abcdef

abcdafc

2

abc

def

样例输出

acdg

acd

0

题意

有n个字符串，每个字符串首尾相连，求这n个字符串的最长公共子序列并输出

思路

将每个首尾相连字符串从0~len-1的每个位置形成的字符串都进行二进制枚举，将所有情况的子序列都用map标记并统计出现的次数，然后枚举map里的元素，将出现n次的子序列存进vector，对vector里的元素进行排序。排序的规则：如果两个子序列长度不同，返回长度较长的子序列，如果相同，返回字典序小的子序列

AC代码

#include<bits/stdc++.h>

using namespace std;

#define line cout<<"------------"<<endl

const int N = 33;

map<string, int>mp,vis;

map<int, string>ms;

vector<string>ve;

int n;

string s,u, str;

bool cmp(string a, string b)

{

    if(a.length() != b.length())

        return a.length() > b.length();

    else

        return a < b;

}

// 求出从pos位置开始的字符串

void change(int pos)

{

    str.clear();

    string temp = s.substr(pos);

    str += temp;

    temp = s.substr(0,pos);

    str += temp;

}

int main()

{

    while(scanf("%d",&n)!=EOF)

    {

        mp.clear();

        ve.clear();

        int nn=n;

        while(nn--)

        {

            vis.clear();

            cin>>s;

            int len=s.length();

            for(int k=0; k<len; k++)

            {

                change(k);

                // 对字符串进行二进制枚举，每个子序列出现的次数用map标记

                for(int i=1;i < (1<<len); i++)

                {

                    for(int j=0; j<len; j++)

                    {

                        if(i >> j & 1)

                           u += str[j];

                    }

                    if(vis[u]==0)

                    {

                       mp[u]++;

                       vis[u]=1;

                    }

                    u.clear();

                }

            }

        }

        int flag=0;

        for(auto i: mp)

        {

            // 找出所有出现次数为n的子序列，存进vector

            if(i.second == n)

            {

                ve.push_back(i.first);

                flag = 1;

            }

        }

        // 对vector进行排序

        sort(ve.begin(),ve.end(),cmp);

        if(flag==0)

            cout<<0<<endl;

        else cout << ve[0] << endl;

    }

    return 0;

}