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

ABCD

BCDFF

BRCD

rose

orchid

Sample Output

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);

        }

    }

}

巴特西

HDOJ1238(string)

Substrings

最新文章

热门文章