POJ - 3080 Blue Jeans

Posted 2020-11-01 duck_lu

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#手动滑稽)

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:

 1 #include <iostream>
 2 #include <cstring>
 3 #include <string>
 4 #include <algorithm>
 5 #include <cstdio>
 6 #include <vector>
 7 #define MAXSIZE 1000100
 8 using namespace std;
 9 
10 vector<string> a;
11 string sub_a = "", res = "";
12 int Next[MAXSIZE];
13 int Len, Len_Res;
14 
15 // int GetNext()
16 // {
17 //     int i = 0, j = Next[0] = -1;
18 //     while (i < Len)
19 //     {
20 //         if (j == -1 || a[i] == a[j])
21 //             Next[++i] = ++j;
22 //         else
23 //             j = Next[j];
24 //     }
25 // }
26 
27 int main(void)
28 {
29     ios::sync_with_stdio(false);
30     int cas;
31     cin >> cas;
32     while (cas--)
33     {
34         int n;
35         cin >> n;
36         for (int i = 0; i < n; ++i)
37         {
38             string temp;
39             cin >> temp;
40             a.push_back(temp);
41         }
42         Len = a[0].size();
43 
44         for (int i = 0; i < Len - 3; ++i)//the origin of str
45         {
46             for (int j = 3; i + j <= Len; ++j)//the len of str
47             {
48                 sub_a = a[0].substr(i, j);
49                 int k;
50                 for (k = 1; k < n; ++k)
51                 {
52                     if (a[k].find(sub_a) == -1)
53                         break;
54                 }
55                 if (k == n)
56                 {
57                     if (sub_a.size() > res.size())
58                         res = sub_a;
59                     else if (sub_a.size() == res.size() && sub_a < res)
60                         res = sub_a;
61                 }
62             }
63         }
64 
65         if (res.size())
66             cout << res << endl;
67         else
68             cout << "no significant commonalities" << endl;
69 
70         //Don‘t forget it!
71         a.clear();
72         res.clear();
73     }
74 
75     return 0;
76 }