HDU-2222 Keywords Search 【AC自动机】

Problem Description

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.

 

Input

First 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.

 

Output

Print how many keywords are contained in the description.

 

Sample Input

1

5

she

he

say

shr

her

yasherhs

 

Sample Output

3

 

---

 

思路：

先把keywords插入Trie,处理出失配指针，再用自动机跑模式串，记录匹配次数时要开一个vis数组，防止多次记录同一结点。

 

 

Code:

  1 #include<iostream>
  2 #include<algorithm>
  3 #include<cstring>
  4 #include<string>
  5 #include<vector>
  6 #include<map>
  7 #include<set>
  8 #include<queue>
  9 char str[1000005], s[55];
 10 
 11 using namespace std;
 12 #define M(a, b) memset(a, b, sizeof(a))
 13 const int maxn = 500000 + 5;
 14 
 15 struct AC_mata {
 16     int f[maxn], last[maxn], val[maxn], ch[maxn][26], sz;
 17     int ans;
 18     bool vis[maxn];
 19 
 20     int idx(char c) {return c - 'a';}
 21 
 22     void init() {
 23         ans = 0;
 24         sz = 0;
 25         M(vis, 0);
 26         M(val, 0); M(f, 0); M(last, 0); M(ch, 0);
 27     }
 28 
 29     void insert(char *s) {
 30         int u = 0, len = strlen(s);
 31         for (int i = 0; i < len; ++i) {
 32             int c = idx(s[i]);
 33             if (!ch[u][c]) ch[u][c] = ++sz;
 34             u = ch[u][c];
 35         }
 36         ++val[u];
 37     }
 38 
 39     void getFail() {
 40         queue<int> q;
 41         f[0] = 0;
 42         for (int i = 0; i < 26; ++i) {
 43             int u = ch[0][i];
 44             if (u) {q.push(u); f[u] = 0; last[u] = 0;}
 45         }
 46         while (!q.empty()) {
 47             int r = q.front(); q.pop();
 48             for (int i = 0; i < 26; ++i) {
 49                 int u = ch[r][i];
 50                 if (!u) {ch[r][i] = ch[f[r]][i]; continue;}
 51                 q.push(u);
 52                 int v = f[r];
 53                 while (v && !ch[v][i]) v = f[v];
 54                 f[u] = ch[v][i];
 55                 last[u] = val[f[u]] ? f[u] : last[f[u]];
 56             }
 57         }
 58     }
 59 
 60     void cnt(int j) {
 61         if (j) {
 62             if (!vis[j]) ans += val[j];
 63             vis[j] = 1;
 64             cnt(last[j]);
 65         }
 66     }
 67 
 68     void find(char *T) {
 69         int j = 0, len = strlen(T);
 70         for (int i = 0; i < len; ++i) {
 71             int c = idx(T[i]);
 72             j = ch[j][c];
 73             if (val[j]) cnt(j);
 74             else if (last[j]) cnt(last[j]);
 75         }
 76     }
 77 
 78 };
 79 
 80 AC_mata ac;
 81 
 82 int main() {
 83     ios::sync_with_stdio(false);
 84     int T, n;
 85     cin >> T;
 86     while (T--) {
 87         ac.init();
 88         cin >> n;
 89         for (int i = 0; i < n; ++i) {
 90             cin >> s;
 91             ac.insert(s);
 92         }
 93         ac.getFail();
 94         cin >> str;
 95         ac.find(str);
 96         cout << ac.ans << endl;
 97     }
 98 
 99     return 0;
100 }

 

转载于:https://www.cnblogs.com/robin1998/p/6551613.html