POJ

题意:给你几个字符串，让你找他们的公共子串(len >= 3, 题上写了，生物也学过)

解题思路:我是拿每种case的第一个串的子串和后面串比较，注意，要按字典序排列，但是大佬说只用找后缀串就行了，至今都没看懂为什么，附上大佬链接，http://blog.sina.com.cn/s/blog_6635898a0100l4fg.html

AC代码:

#include <iostream>
#include <cstring>
#include <string>
#include <algorithm>
#include <cstdio>
#include <vector>
#define MAXSIZE 1000100
using namespace std;

vector<string> a;
string sub_a = "", res = "";
int Next[MAXSIZE];
int Len, Len_Res;

// int GetNext()
// {
//     int i = 0, j = Next[0] = -1;
//     while (i < Len)
//     {
//         if (j == -1 || a[i] == a[j])
//             Next[++i] = ++j;
//         else
//             j = Next[j];
//     }
// }

int main(void)
{
    ios::sync_with_stdio(false);
    int cas;
    cin >> cas;
    while (cas--)
    {
        int n;
        cin >> n;
        for (int i = 0; i < n; ++i)
        {
            string temp;
            cin >> temp;
            a.push_back(temp);
        }
        Len = a[0].size();

        for (int i = 0; i < Len - 3; ++i)//the origin of str
        {
            for (int j = 3; i + j <= Len; ++j)//the len of str
            {
                sub_a = a[0].substr(i, j);
                int k;
                for (k = 1; k < n; ++k)
                {
                    if (a[k].find(sub_a) == -1)
                        break;
                }
                if (k == n)
                {
                    if (sub_a.size() > res.size())
                        res = sub_a;
                    else if (sub_a.size() == res.size() && sub_a < res)
                        res = sub_a;
                }
            }
        }

        if (res.size())
            cout << res << endl;
        else
            cout << "no significant commonalities" << endl;

        //Don't forget it!
        a.clear();
        res.clear();
    }

    return 0;
}