時間限制:1000ms

單點時限:1000ms

記憶體限制: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.

輸入

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.

輸出

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

樣例輸入

2
abcdefg
zaxcdkgb
5
abcdef
kedajceu
adbac
abcdef
abcdafc
2
abc
def

樣例輸出

acdg
acd
0

資料量非常小，根本用不到什麼演算法，暴搜出每一個字串的所有子序列，每一個字串的所有子序列加入一個set裡，最後求n個set的交集中最長的那個字串字典序最小的即可

#include <bits/stdc++.h>
using namespace std;
set<string> st[11];
string t,ans;
int len;
void dfs(int i,int cur)
{
    if(cur == len) {
        if(ans == "") return;
        st[i].insert(ans);
        int _len = ans.length();
        for(int j = 1; j < _len; j++) {
            st[i].insert(ans.substr(j,len - j) + ans.substr(0,j));
        }
        return;
    }
    for(int j = 0; j <= 1; j++) {
        if(j == 0) ans.push_back(t[cur]);
        dfs(i,cur + 1);
        if(j == 0) ans.erase(ans.end() - 1);
    }
}
int main(void)
{
    int n;
    while(scanf("%d",&n) != EOF) {
        for(int i = 0; i < n; i++) {
            st[i].clear();
        }
        for(int i = 0; i < n; i++) {
            cin >> t;
            len = t.length();
            dfs(i,0);
        }
        set<string>::iterator it;
        string ans = "0";
        int flag;
        for(it = st[0].begin(); it != st[0].end(); it++) {
            t = *it;
            flag = 0;
            for(int i = 1; i < n; i++) {
                if(!st[i].count(t)) {
                    flag = 1;
                    break;
                }
            }
            if(flag == 0) {
                if(ans == "0") {
                    ans = t;
                }
                else if(t.length() > ans.length()) {
                    ans = t;
                }
                else if(t.length() == ans.length() && ans > t) {
                    ans = t;
                }
            }
        }
        cout << ans << endl;
    }
    return 0;
}