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.

The first line of the input 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.

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

2

3

ABCD

BCDFF

BRCD

2

rose

orchid

2

2 

题目大意：
给你n个字符串  要你找这n个字符串中最大的相同字串（字符串可以逆置）

分析：
先找到最小的母串  然后这个母串的所有字串从大到小 都和n个串循环一遍  只要找到符合的答案就跳出

#include<stdio.h>

#include<string.h>

#include<stdlib.h>

#include<iostream>

#include<algorithm>

#include<math.h>

#include<queue>

#define N 220

#define INF 0xffffffff

using namespace std;

int main()

{

    int T,n;

    char str[N][N],s1[N],s2[N];

    scanf("%d",&T);

    while(T--)

    {

        scanf("%d",&n);

        int Min=INF,b;

        for(int i=; i<n; i++)

        {

            scanf("%s",str[i]);

            if(strlen(str[i])<Min)

            {

                Min=strlen(str[i]);

                b=i;

            }

        }

        int len=Min,flag;

        while(len)

        {

            for(int i=; i<=Min-len; i++)

            {

                flag=;

                strncpy(s1,str[b]+i,len);

                int k=;

                for(int i=len-; i>=; i--)

                {

                    s2[k++]=s1[i];

                }

                s1[len]=s2[len]='\0';

                for(int i=; i<n; i++)

                {

                    if(strstr(str[i],s1)==NULL && strstr(str[i],s2)==NULL)

                    {

                        flag=;

                        break;

                    }

                }

                if(flag==)

                {

                    break;

                }

            }

            if(flag==)

                break;

            len--;

        }

        printf("%d\n",len);

    }

    return ;

}