I - Substrings

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. 

InputThe 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. 
OutputThere should be one line per test case containing the length of the largest string found. 
Sample Input

2
3
ABCD
BCDFF
BRCD
2
rose
orchid

Sample Output

2

2

题目大意：

题目的意思就是求最小的字串长度，字串可以是反串，也可以是本身的字串，也就是本身可以不是字串，但是他的反串也可以是他的子串。那么长度依然可以是这样长。

思路就是：找到一个最小的长度，作为一个寻找串，求的他的个个字串，之后就是比较，其他长度的串，可以不要去寻找了，因为找的字串，每个都要有。所以我们只要找一遍就可以了。选择的自然就是最小长度的串。减少时间！

代码如下：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string.h>
using namespace std;

int main()
{
    int n,t;
    scanf("%d",&n);
    while(n--)
    {
        int m;
        scanf("%d",&m);
        string fre[105];
        int len = 99999;
        int o;
        for(int i=0; i<m; i++)
        {
            cin>>fre[i];
            if(len>fre[i].length())
            {
                len = fre[i].length();
                o=i;
            }
        }
        int max = 0;
        for(int j=len; j>=1; j--)
        {
            for(int k=0; k<=len-j; k++)
            {
            	//从k位置开始，截取到j长度的位置。 
            	string ans = fre[o].substr(k,j);
            	string ans1 = ans;
                reverse(ans1.begin(),ans1.end());
                for( t=0;t<m;t++)
                {
                    if(fre[t].find(ans1)==-1 && fre[t].find(ans)==-1)
                    {
                        break;
                    }
                }
                if(t==m)
                {
                    if(max<ans1.length())
                    {
                        max = ans1.length();
                        break;
                    }
                }
            }
       }
        printf("%d\n",max);
    }
    return 0;
}