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

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原文链接：http://www.cnblogs.com/l609929321/p/7538043.html

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#java

本文介绍了一个关于字符串处理的问题——寻找能够支配所有其他字符串的支配者。通过输入一系列字符串，程序将判断是否存在一个字符串作为支配者，即该字符串包含所有其他字符串作为子串。若存在，则输出该支配者；否则输出“No”。示例展示了不同情况下支配者的存在与否。

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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

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

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;
}