Substrings（hd1238）

Substrings

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8183    Accepted Submission(s):
3752

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

找到最短字符串，比如其长度为5，先取5的子串，再取4的子串...每次遍历其他字符串中是否含有其逆串或顺串

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 using namespace std;
 /* 原型：char * strncpy(char *dest, char *src, size_t n); 　　功能：将字符串src中最多n个字符复制到字符数组dest中(它并不像strcpy一样只有遇到NULL才停止复制，而是多了一个条件停止，就是说如果复制到第n个字符还未遇到NULL，也一样停止），返回指向dest的指针。*/
 int main()
 {
     int N,i,j,num;
     freopen("in.txt","r",stdin);
     cin>>N;
     while(N--)
     {
         char string[][],pos[],inv[],str[];
         int min_str=,index,len,flag=;
         cin>>num;
         for(i=;i<num;i++)
         {
             cin>>string[i];//相当于把\n换成'\0';
             if(strlen(string[i])<min_str)
                 min_str=strlen(string[i]);
             index=i;
         }//输入字符串，并找到短字符串
         len=min_str;
         strcpy(str,string[index]);
         while(len>)
         {
             for(i=;i<=min_str-len;i++)//对于len长的子串可以取多少种；从最长的开始取
             {
                 flag=;//假设该子串符合
                 strncpy(pos,str+i,len);//并不会把'\0'复制进去，自己加进去
                 for(j=;j<len;j++)
                     inv[j]=pos[len-j-];//求出逆向子串；
                 inv[len]=pos[len]='\0';
                 for(j=;j<num;j++)
                 {
                     if(strstr(string[j],inv)==NULL&&strstr(string[j],pos)==NULL)
                     {
                         flag=;
                         break;
                     }
                 }
                 if(flag)
                     break;
             }
             if(flag)
                 break;
             len--;
         }
         cout<<len<<endl;
     }
 }