2017 ACM/ICPC Asia Regional Qingdao Online 1003 The Dominator of Strings hdu 6208

The Dominator of Strings

Time Limit: 3000/3000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 0    Accepted Submission(s): 0

Problem Description

Here you have a set of strings. A dominator is a string of the set dominating all strings else. The string S is dominated by T if S is a substring of T.

 

 

Input

The input contains several test cases and the first line provides the total number of cases.
For each test case, the first line contains an integer N indicating the size of the set.
Each of the following N lines describes a string of the set in lowercase.
The total length of strings in each case has the limit of 100000.
The limit is 30MB for the input file.

 

 

Output

For each test case, output a dominator if exist, or No if not.

 

 

Sample Input

3 10 you better worse richer poorer sickness health death faithfulness youbemyweddedwifebetterworsericherpoorersicknesshealthtilldeathdouspartandpledgeyoumyfaithfulness 5 abc cde abcde abcde bcde 3 aaaaa aaaab aaaac

 

 

Sample Output

youbemyweddedwifebetterworsericherpoorersicknesshealthtilldeathdouspartandpledgeyoumyfaithfulness abcde No

 

 

Source

输入输出测试

 

#include <bits/stdc++.h>
#define IO ios::sync_with_stdio(false); cin.tie(0)
using namespace std ;
vector<string> dic ;
string str ,tp ;
int main()
{
    IO;
    int t,n;

    cin >> t ;
    while(t--)
    {
        int n , len = -1 , buf ;
        cin >> n ;
        dic.clear() ;
        for(int i = 0 ; i < n ; i++ )
        {
            cin >> tp ;
            dic.push_back(tp) ;
            int x = tp.length();
            if( x > len )
            {
                len = x ;
                str = tp ;
            }
        }

        bool flag = true ;
        for( int i = 0 ; i < n ; i++ )
        {

            if( str.find(dic[i]) == string::npos )
            {
                flag = false ;
                break ;
            }
        }
        if( flag )
        {
            cout << str << endl ;
        }
        else
        {
            cout << "No" <<endl ;
        }
    }
    return 0;
}