poj1226--Substrings

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 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>

int main()

{

       int i,j,k,n,t,len,flag,minlen;

       char str[110][110],a[110],b[110],c[110];

       scanf("%d",&t);

       while(t--)

       {

              scanf("%d",&n);

              minlen=220;

              flag=0;

              for(i=0;i<n;i++)     //因为公共子串的长度不会长于最短的那个字符串，所以用minlen记录n个字符串的最短长度，并把那个最短的字符串存在c中

              {

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

                     len=strlen(str[i]);

                     if(minlen>len)

                     {

                            minlen=len;

                            strcpy(c,str[i]);

                     }

              }

              for(i=minlen;i>0;i--)//i为子串的长度

              {

                     for(j=0;j<=minlen-i;j++)//当子串的长度为i时，存在minlen-i种可能的字串。例如ABCD,当子串长度为4时，子串只有一种，即ABCD；当子串的长度为3时，有ABC,BCD两种；当子串长度为2时，有AB,BC,CD三种子串

                     {

                            strncpy(a,&c[j],i);//strncpy(*str1,*str2,n)就是从str2中copy i个字符赋给str1；

                            a[i]='\0';

                            strcpy(b,a);

                            b[i]='\0';

                            strrev(b);

                            for(k=0;k<n;k++)

                            {

                                   if(strstr(str[k],a)==NULL&&strstr(str[k],b)==NULL)//strstr(s1,s2)如果s1,是s2的子串，返回s1第一次出现的地址，否则返回NULL;

                                          break;

                            }

                            if(k==n)

                            {

                                   flag=i;

                                   break;

                            }

                     }

                     if(flag)

                            break;

              }

              printf("%d\n",flag);

       }

       return 0;

}

 

 

Language:Default Substrings Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 7402 Accepted: 2475 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 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