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[100][103],pos[103],inv[103],str[103];
        int min_str=100,index,len,flag=0;
        cin>>num;
        for(i=0;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>0)
        {
            for(i=0;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=1;//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=0;j<len;j++)
                    inv[j]=pos[len-j-1];//求出逆向子串；
                inv[len]=pos[len]='\0';
                for(j=0;j<num;j++)
                {
                    if(strstr(string[j],inv)==NULL&&strstr(string[j],pos)==NULL)
                    {
                        flag=0;
                        break;
                    }
                }
                if(flag)
                    break;
            }
            if(flag)
                break;
            len--;
        }
        cout<<len<<endl;
    }
}