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Substrings

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11034    Accepted Submission(s): 5310


Problem Description

You are given a number of case-sensitive strings of alphabetic characters, find the largest string X, such that either X, or its inverse can be found as a substring of any of the given strings.

 


Input

The first line of the input file contains a single integer t (1 <= t <= 10), the number of test cases, followed by the input data for each test case. The first line of each test case contains a single integer n (1 <= n <= 100), the number of given strings, followed by n lines, each representing one string of minimum length 1 and maximum length 100. There is no extra white space before and after a string. 

 


Output

There should be one line per test case containing the length of the largest string found.

 


Sample Input

2

3

ABCD

BCDF

F BRCD

2

rose

orchid

 


Sample Output

2 2


找到最短的串，暴力所有字串，去匹配。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
using namespace std;
int main()
{
    int ca,n,f;
    scanf("%d",&ca);
    string s[105];
    while(ca--)
    {
        int mma=1000;
        int ma=0;
        scanf("%d",&n);
        for(int i=0;i<n;i++)
        {
            cin>>s[i];
            if(s[i].size()<mma)
            {
                mma=s[i].size();
                f=i;
            }
        }
        int le=s[f].size();
        for(int i=0;i<le;i++)
        {
            for(int j=0;j<le;j++)
            {
                string s1,s2;
                s1=s[f].substr(i,j);
                s2=s1;
                reverse(s2.begin(),s2.end());
                int flag=1;
                for(int k=0;k<n;k++)
                {
                    if(s[k].find(s1,0)==-1&&s[k].find(s2,0)==-1)
                    {
                        flag=0;
                        break;
                    }
                }
                if(flag&&s1.size()>ma)
                    ma=s1.size();
            }
        }
        cout<<ma<<endl;
    }
    return 0;
}