Substrings

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7205 Accepted Submission(s):
3255

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

这个找字串的问题，题目大概意思就是找出所有字符串中共同拥有的一个子串，

该子串（正、逆字符）是任何一个母串的子串，求该子串的最长长度。

想到用STRING里的成员函数和STL的reverse反转函数，

思路:

先找出最短的母串，即该符合要求的子串肯定在这个母串中，即在从长到短

从最短母串中取子串，在子串正反去查看是否符合要求。

说实话今天又学到了一些知识，我表示这C++的很多函数真shi 强大啊。

ps:http://acm.hdu.edu.cn/showproblem.php?pid=1238

#include<iostream>

#include<string>

#include<algorithm>//STL reverse函数的头文件，reverse反转函数，

using namespace std;

int main()

{

    int cas,len,sub,maxn;

    int n,k,i,j;

    string s[];

    cin>>cas;

    while(cas--)

    {

        cin>>n;

        len=;

        sub=;

        for(i=; i<n; i++)

        {

            cin>>s[i];

            if(len>s[i].size())//找最小 的母串

            {

                len=s[i].size();

                sub=i;

            }

        }

        maxn=;

        for(i=s[sub].size(); i>; i--) //从最小的母串开始从长到短找子串，

        {

            for(j=; j<s[sub].size()-i+; j++) //长度为i的子串在母串中找

            {

                string s1,s2;//s1为子串正 ，s2为子串反

                s1=s[sub].substr(j,i);//去j开始i长度是字符

                s2=s1;

                reverse(s2.begin(),s2.end());//反串

                for( k=; k<n; k++)

                {

                    if(s[k].find(s1,)==-&&s[k].find(s2,)==-) //当正反子串在母串中都未发现时即跳出

                        break;

                }

                if(k==n&&maxn<s1.size())

                    maxn=s1.size();

            }

        }

        cout<<maxn<<endl;

    }

    return ;

}

巴特西

Substrings（hdu1238）字符串匹配

Substrings

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