In the modern time, Search engine came into the life of everybody like Google, Baidu, etc.
Wiskey also wants to bring this feature to his image retrieval system.
Every image have a long description, when users type some keywords to find the image, the system will match the keywords with description of image and show the image which the most keywords be matched.
To simplify the problem, giving you a description of image, and some keywords, you should tell me how many keywords will be match.
Wiskey also wants to bring this feature to his image retrieval system.
Every image have a long description, when users type some keywords to find the image, the system will match the keywords with description of image and show the image which the most keywords be matched.
To simplify the problem, giving you a description of image, and some keywords, you should tell me how many keywords will be match.
InputFirst line will contain one integer means how many cases will follow by.
Each case will contain two integers N means the number of keywords and N keywords follow. (N <= 10000)
Each keyword will only contains characters 'a'-'z', and the length will be not longer than 50.
The last line is the description, and the length will be not longer than 1000000.
OutputPrint how many keywords are contained in the description.Sample Input
1
5
she
he
say
shr
her
yasherhs
Sample Output
3
题意：
T组输入，对于每组数据有n个模式串，最后输入一个待求串。让你求这个待求串中有多少种模式串
注意：这个模式串可能会重复，重复的话要算做多个，不能算作一个
代码：
1 /*
2 代码中：
3 叶节点：代表此节点下没有子节点
4 根节点：就是树的根
5 子节点：就是这个节点的直接相连的节点(直系节点)
6
7 此代码：
8 用题目所给模式串构成一颗字典树
9 然后找出来给出的待求串中有多少种模式串(如果模式串中有重复的，那么也算作多个，不算一个)
10
11 */
12 #include<stdio.h>
13 #include<iostream>
14 #include<string.h>
15 #include<algorithm>
16 #include<queue>
17 using namespace std;
18 const int maxn=5e5+10;
19 const int N=26;
20 typedef long long ll;
21 struct Trie
22 {
23 int next[maxn][N],fail[maxn],ends[maxn];
24 int root,L;
25 int New_node() //创建一个新节点
26 {
27 for(int i=0;i<N;++i)
28 {
29 next[L][i]=-1; //每次初始化新节点
30 }
31 ends[L++]=0; //每次初始化新节点
32 return L-1;
33 }
34 void init() //创建根节点
35 {
36 L=0;
37 root=New_node();
38 }
39 void inserts(char s[]) //往字典树里面插入新字符串
40 {
41 int len=strlen(s);
42 int now=root;
43 for(int i=0;i<len;++i)
44 {
45 if(next[now][s[i]-'a']==-1)
46 next[now][s[i]-'a']=New_node();
47 now=next[now][s[i]-'a'];
48 }
49 ends[now]++; //字典树的叶节点要标记，如果走到这个位置就代表这种模式串出现过
50 }
51 void build()
52 {
53 queue<int>r;
54 fail[root]=root;
55 for(int i=0;i<N;++i) //先给字典树根节点下的N个子节点初始化一下
56 {
57 if(next[root][i]==-1) //代表字典树上都没有i这个分支，对于这些节点的下一级就直接指向根节点就完了
58 {
59 next[root][i]=root;
60 }
61 else //字典树上有i这个分支的，就要改变它的fail(即，失败指针)的值
62 {
63 fail[next[root][i]]=root;
64 r.push(next[root][i]);
65 }
66 }
67 while(!r.empty())
68 {
69 int now=r.front();
70 r.pop();
71 for(int i=0;i<N;++i)
72 {
73 if(next[now][i]==-1)
74 {
75 next[now][i]=next[fail[now]][i];
76 //代表now在叶节点处，此时往下一个节点走就直接通过now节点失败指针进行跳转
77 }
78 else
79 {
80 fail[next[now][i]]=next[fail[now]][i]; //此节点的子节点的失败指针就是此节点失败指针对应字符位置
81 r.push(next[now][i]);
82 }
83 }
84 }
85 }
86 int query(char s[])
87 {
88 int len=strlen(s);
89 int now=root;
90 int res=0;
91 for(int i=0;i<len;++i)
92 {
93 now=next[now][s[i]-'a']; //这个now就保证了我们找的肯定是这个s字符串中的一部分
94 int temp=now; //这就是一直找，找到了就加1，找不到就通过失败指针看其他地方能不能找到
95 while(temp!=root) //尝试这个节点的失败指针能不能找到模式串
96 {
97 res+=ends[temp]; //因为我们要找出来s串中包含几种模式串，所以要有赋值0操作
98 ends[temp]=0;
99 temp=fail[temp];
100 }
101 }
102 return res;
103 }
104 };
105 char s[1000010];
106 Trie ac;
107 int main()
108 {
109 int t;
110 scanf("%d",&t);
111 while(t--)
112 {
113 int n;
114 scanf("%d",&n);
115 ac.init();
116 while(n--)
117 {
118 scanf("%s",s); //这个就是模式串
119 ac.inserts(s);
120 }
121 ac.build();
122 scanf("%s",s); //这个是待求串
123 printf("%d\n",ac.query(s));
124 }
125 return 0;
126 }