Substrings（hdu1238）字符串匹配

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

2

3

ABCD

BCDFF

BRCD

2

rose

orchid

 

 

Sample Output

2

2

 

 

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[102];
    cin>>cas;
    while(cas--)
    {
        cin>>n;
        len=1000;
        sub=0;
        for(i=0; i<n; i++)
        {
            cin>>s[i];
            if(len>s[i].size())//找最小 的母串
            {
                len=s[i].size();
                sub=i;
            }
        }
        maxn=0;
        for(i=s[sub].size(); i>0; i--) //从最小的母串开始从长到短找子串，
        {
            for(j=0; j<s[sub].size()-i+1; 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=0; k<n; k++)
                {
                    if(s[k].find(s1,0)==-1&&s[k].find(s2,0)==-1) //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 0;
}