poj1226

Substrings

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 8467 Accepted: 2890

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

23ABCDBCDFFBRCD2roseorchid

Sample Output

22 这个题目式模式匹配的一个问题，就是在多个字符串中找最大子串。。。。

# 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s2,是s1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;}