HDU_1238——最大子串搜索

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 BCDFF BRCD 2 rose orchid

 

 

Sample Output

2 2

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <windows.h>
void sort(char (*str)[101]);  //给字符串序列按长度排序 
int fun(char (*str)[101]);   //求发现的最大的子串的长度 
//char* in_turn(char* s);    //字符串逆序 

int main()
{
   int t,n;
   scanf("%d",&t);
   while(t--)   //提前判断范围 
      {
         int i=0;
         char str[100][101]={0};
         scanf("%d",&n);
         while(i<n)
            {               
               scanf("%s",str[i++]);   //gets会接收一个换行符 
            }
         sort(str);
         printf("%d\n",fun(str));  
      } 
   system("pause>nul");
   return 0;
}

int fun(char (*str)[101])
{
   int max=0,i,j,k;
   char s[101];
   for(k=0;str[0][k];k++)
   {
      for(i=1;i<=strlen(str[0])-k;i++)
         {
           if(max>i) continue;
           int flag=1;
           strcpy(s,str[0]+k);//i-k+1
           s[i]='\0';        
           for(j=1;str[j][0];j++)
               {
               if(strstr(str[j],s)==NULL && strstr(str[j],strrev(s))==NULL)  //返回一个指向s1中第一次出现s2中字符串的位置，没有返回NULL  
                  {flag=0;break;}                                            //strrev()逆序函数 
               }
            if(flag==1)   
               max=i;
         }
   }
   return max;
}
/*
char* in_turn(char* str)
{
   char _str[101];
   int i;
   for(i=0;str[i];i++)
      {
         _str[strlen(str)-1-i] = str[i];   
      }
   _str[strlen(str)]='\0';
   return _str;
}
*/
void sort(char (*str)[101])   //一个指向100个char类型的指针 
{
   int i,j;
   char temp[100];
   for(i=0;str[i+1][0]!='\0';i++)
      for(j=i+1;str[j][0]!='\0';j++)
         {
            if(strlen(str[i]) > strlen(str[j]))
               {
                  strcpy(temp,str[i]);
                  strcpy(str[i],str[j]);
                  strcpy(str[j],temp);
               }   
         }
}