1.链接地址：

http://bailian.openjudge.cn/practice/1226/

http://poj.org/problem?id=1226

2.题目：

总时间限制:

1000ms

内存限制:

65536kB

描述

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.

输入

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.

输出

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

样例输入

23ABCDBCDFFBRCD2roseorchid

样例输出

22

来源

Tehran 2002 Preliminery

3.思路：

 

4.代码：

1 #include 
     
       2 #include 
      
        3 #include 
       
         4 #include 
        
          5 #include 
         
           6 using namespace std; 7 const int NUM = 100; 8 char strs[NUM][NUM + 1]; 9 int main()10 {11     //freopen("F:\\input.txt","r",stdin);12     int i,j,k;13 14     int t;15     cin>>t;16     17     int n,length;18     while(t--)19     {20         cin>>n;21         cin.get();22         23         for(i = 0; i < n; i++)24         {25             scanf("%s",strs[i]);26         }27         28         for(i = 0; i < n; i++)29 30         length = strlen(strs[0]);31         char substr[NUM + 1],substr2[NUM + 1];32         int res = 0;33         for(i = 1; i <= length; i++)34         {35             for(j = 0; (j+i-1) < length; j++)36             {37                 strncpy(substr,&strs[0][j],i);38                 substr[i] = '\0';39                 strcpy(substr2,substr);40                 41                 for(k = 0; k < (i+1)/2; k++)42                 {43                     char tmp = substr2[k];44                     substr2[k] = substr2[i-1-k];45                     substr2[i-1-k] = tmp;46                 }47                 48                 49                 //cout<<"substr="<
          
           <<",substr2="<
           
            <
            
             = n ) 55 {56 res = i;57 break;58 }59 }60 }61 62 cout<
             
              <