HDOJ1238(string)

Substrings

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9695    Accepted Submission(s): 4602

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

暴力枚举:时间复杂度o(10^6)。

import java.util.Arrays;
import java.util.Scanner;
import java.util.Comparator;
class Comp implements Comparator<String>{
    public int compare(String s1,String s2)
    {
        return s1.length()-s2.length();
    }
}
public class Main{
    static Scanner cin=new Scanner(System.in);
    static final int MAXN=105;
    static int n;
    static String[] s = new String[MAXN];
    static String reverse(String s)
    {
        String rev="";
        int len=s.length();
        for(int i=0;i<len;i++)
        {
            rev=s.charAt(i)+rev;
        }
        return rev;
    }
    public static void main(String[] args){
        int T=cin.nextInt();
        while(T--!=0)
        {
            int n;
            n=cin.nextInt();
            for(int i=0;i<n;i++)
            {
                s[i]=cin.next();
            }
            Arrays.sort(s,0,n,new Comp());
            int len=s[0].length();
            label:
            while(len>0)
            {
                String ss;
                String rs;
                int s0len=s[0].length();
                for(int start=0;start+len<=s0len;start++)
                {
                    ss=s[0].substring(start, start+len);
                    rs=reverse(ss);
                    int mark=0;
                    for(int j=1;j<n;j++)
                    {
                        if(s[j].indexOf(ss)!=-1||s[j].indexOf(rs)!=-1)
                        {
                            mark++;
                        }
                    }
                    if(mark==n-1)
                    {
                        break label;
                    }
                }
                len--;
            }
            System.out.println(len);
        }
    }
}