3           //case1案例库
aaa
aaa
baa
2           //给定字符串
aa
ba
1           //case2 案例库
a
1            //给定字符串
a输出
3       //aa是案例库1中三个字符串的子串，个数为3
1       //ba是案例库1中第三个字符串的子串，故个数为1
1       //a是案例库2中的字串，个数为1

 1 #include<iostream>
 2 #include<string>
 3 #include<vector>
 4
 5 using namespace std;
 6
 7 bool isMatch(const string &dict, const string &example);
 8 vector<int> getNext(const string &needle, int m, vector<int> next);
 9
10 int main()
11 {
12     while (cin)        //此处是否正确？
13     {
14         int N;
15         cin >> N;
16         vector<string> dict(N, "");
17         for (int i = 0;i<N;i++)
18         {
19             cin >> dict[i];
20         }
21
22         int M;
23         cin >> M;
24         vector<string> example(M, "");
25         for (int i = 0;i<M;i++)
26         {
27             cin >> example[i];
28         }
29
30         for (int i = 0;i<M;i++)
31         {
32             int count = 0;
33             for (int j = 0;j<N;j++)
34             {
35                 if (isMatch(dict[j], example[i]))
36                 {
37                     count++;
38                 }
39             }
40             cout << count << endl;        //输出
41         }
42     }
43     return 0;
44 }
45
46 //KMP匹配算法
47 bool isMatch(const string &dict, const string &example)
48 {
49     int n = dict.size(), i = 0;
50     int m = example.size(), j = 0;
51
52     if (m>n) return false;
53     if (n == 0 || m == 0)  return true;
54
55     vector<int> next(m);
56     next = getNext(example, m, next);
57
58     while (j<m&&i<n)
59     {
60         if ((0>j) || dict[i] == example[j])
61         {
62             i++;
63             j++;
64         }
65         else
66         {
67             j = next[j];
68         }
69     }
70     return j == m ? true : false;
71 }
72
73 vector<int> getNext(const string &needle, int m, vector<int> next)
74 {
75     int t = -1;
76     next[0] = -1;
77     int j = 0;
78     while (j<m - 1)
79     {
80         if (0>t || needle[j] == needle[t])
81         {
82             j++;
83             t++;
84             next[j] = (needle[j] != needle[t] ? t : next[t]);
85         }
86         else
87             t = next[t];
88     }
89     return next;
90 }