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)[]);  //给字符串序列按长度排序
 int fun(char (*str)[]);   //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=;
          char str[][]={};
          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 ;
 }

 int fun(char (*str)[])
 {
    int max=,i,j,k;
    char s[];
    for(k=;str[][k];k++)
    {
       for(i=;i<=strlen(str[])-k;i++)
          {
            if(max>i) continue;
            int flag=;
            strcpy(s,str[]+k);//i-k+1
            s[i]='\0';
            for(j=;str[j][];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=;break;}                                            //strrev()逆序函数
                }
             if(flag==)
                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)[])   //一个指向100个char类型的指针
 {
    int i,j;
    char temp[];
    for(i=;str[i+][]!='\0';i++)
       for(j=i+;str[j][]!='\0';j++)
          {
             if(strlen(str[i]) > strlen(str[j]))
                {
                   strcpy(temp,str[i]);
                   strcpy(str[i],str[j]);
                   strcpy(str[j],temp);
                }
          }
 }