HDU-6208 最大串的子串

最新推荐文章于 2019-07-27 11:12:09 发布

luer9

最新推荐文章于 2019-07-27 11:12:09 发布

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分类专栏： vector

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本文介绍了一种算法，用于解决给定字符串集合中寻找支配字符串的问题。支配字符串是指该字符串包含集合中所有其他字符串作为其子串。文章通过示例说明了如何实现这一功能，并提供了一个C++代码示例。

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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
代码如下：
#include<iostream>
#include<string>
#include<vector>
using namespace std;

int main()
{
	ios::sync_with_stdio(false);  //存在大量输入输出时需关闭IO同步或使用scanf与printf 
	
	int n,t,flag;
	string str,ans;
	
	cin>>t;
	
	while(t--)
	{
		vector<string>vec;  //定义一个string容器 
		vector<string>::iterator it;
		
		cin>>n;
		
		flag=1;
		ans.clear();  //多组数据，使用前需清空 
		
		for(int i=1;i<=n;i++)
		{
			cin>>str;
			vec.push_back(str);
			
			if(ans.length()<str.length())  //每次比较长度，找到最大子串 
				ans=str;
		}
		
		for(it=vec.begin();it!=vec.end();it++)
		{
			if(ans.find(*it)==ans.npos)  //判断当前字符串是否为最大串的子串 
			{
				flag=0;
				break;
			}
		}
		
		if(flag)
			cout<<ans<<endl;
		else
			cout<<"No"<<endl;
	}
	return 0;
}