hdu 1238 Substrings

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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

23ABCDBCDFFBRCD2roseorchid

Sample Output

22
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第一道搜索题
枚举长度最短的字符串的所有字串，然后与其它字符串比较
一旦比较失败，就跳出；接着循环，枚举下一个，
#include<stdio.h>#include<string.h>#include<stdlib.h>char str[110][110];int main(){ int t,n,i,len,j,flag; int min,max,p,x,y,k,q; char a[102],b[102]; char c[102]; scanf("%d",&t); while(t--) {  scanf("%d",&n);  memset(str,0,sizeof(str));  min =110;  for(i=0;i<n;i++)  {   scanf("%s",str[i]);   len = strlen(str[i]);   if(len<min)   {    min = len;    p = i;   }  }   max=0;  for(i=0;i<min;i++)  {   for(j=1;j<=min-i;j++)// 此循环，控制以某一个字母开头的所有字串，下面的查找循环中，如果，没有找到以此字母开头的字串，就跳出；
                       //枚举以下一个字母开头的所有字串   {    strncpy(a,str[p]+i,j);//枚举str[p]的字串；    a[j]='\0';    //printf("a:%s\n",a);    //b = strrev(a);    x=0;q=0;    for(k=j-1;k>=0;k--)       b[x++]=a[k];    b[x]='\0';    //printf("b:%s\n",b);    flag=1;    for(y=0;y<n;y++) //这里是与其它字符串比较，一旦查找失败，就跳出    {     if(y!=p)     {      if(strstr(str[y],a)==NULL&&strstr(str[y],b)==NULL)      { flag=0 ;break; }     }    }    if(!flag)// 以某一个字母开头的第一个子串查找失败，那么就没有必要查找以该字母开头的其它子串，所以这里跳出第二个for循环     break;    if(max<j)//这里负责找到长度最大的字串    {     max=j;     //for(x=0;x<j;x++) // 加上这几句，就能求出 最长公共字串；     // c[q++] = a[x];     //c[x]='\0';    }   }  }  printf("%d\n",max); } return 0;}

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