HDU 2222 Keywords Search（AC自动机/模板题）

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Keywords Search

Time Limit: 2000/1000 MS (Java/Others)            Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 63941             Accepted Submission(s): 21264

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

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#include<bits/stdc++.h> using namespace std;   struct Trie{     int cnt;     Trie* next[26];     Trie* behind; };     int main() {     int T;     scanf("%d",&T);     while (T--)     {         int N;         char str[60];         Trie* root = new Trie;         root->cnt = -1;         root->behind = NULL;         for (int i = 0;i < 26;i++)  root->next[i] = NULL;         scanf("%d",&N);         while (N--)         {             scanf("%s",str);             Trie* p = root;             int i = 0;             while (str[i])             {                 int j = str[i] - 'a';                 if (!p->next[j])                 {                     Trie* q = new Trie;                     q->cnt = 0;                     q->behind = NULL;                     for (int k = 0;k < 26;k++)  q->next[k] = NULL;                     p->next[j] = q;                 }                 p = p->next[j];                 i++;             }             p->cnt++;         }         queue<Trie*>que;         root->behind = root;         for (int i = 0;i < 26;i++)         {             if (!root->next[i])             {                 root->next[i] = root;             }             else             {                 root->next[i]->behind = root;                 que.push(root->next[i]);             }         }         while (!que.empty())         {             Trie* p = que.front();             Trie* q = p->behind;             que.pop();             for (int i = 0;i < 26;i++)             {                 if (!p->next[i])                 {                     p->next[i] = q->next[i];                 }                 else                 {                     p->next[i]->behind = q->next[i];                     que.push(p->next[i]);                 }             }         }         char article[1000005];         Trie* p = root;         scanf("%s",article);         int res = 0;         int i = 0;         while (article[i])         {             int j = article[i] - 'a';             p = p->next[j];             for (Trie* tmp = p;tmp != root && tmp->cnt !=-1;tmp = tmp->behind)             {                 res += tmp->cnt;                 tmp->cnt = -1;    //计算完毕，置-1表示已经计算过             }             i++;         }         printf("%d\n",res);     }     return 0; }

　　

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#include<bits/stdc++.h> using namespace std; const int N = 500005;   struct AC_Automan{     int next[N][26];     int fail[N];     int cnt[N];     int root,tot;     int newnode()     {         for (int i = 0;i < 26;i++)  next[tot][i] = -1;             fail[tot] = -1;         cnt[tot] = 0;         return tot++;     }     void clear()     {         tot = 0;         root = newnode();     }     void insert(const string &s)     {         int p = root;         for (int i = 0,len = s.length();i < len;i++)         {             if (next[p][s[i]-'a'] == -1)    next[p][s[i]-'a'] = newnode();             p = next[p][s[i]-'a'];         }         cnt[p]++;     }     void build()     {         queue<int>Q;         Q.push(root);         while (!Q.empty())         {             int p = Q.front();Q.pop();             for (int i = 0;i < 26;i++)             {                 if (~next[p][i])                 {                     if (p == root)  fail[next[p][i]] = root;                     else    fail[next[p][i]] = next[fail[p]][i];                     Q.push(next[p][i]);                 }                 else                 {                     if (p == root)  next[p][i] = root;                     else    next[p][i] = next[fail[p]][i];                 }             }         }     }     int solve(const string &t)     {         int p = root,ans = 0;         for (int i = 0,len = t.length();i < len;i++)         {             p = next[p][t[i]-'a'];             for (int tmp = p;tmp != root && cnt[tmp] != -1;tmp = fail[tmp])             {                 ans += cnt[tmp];                 cnt[tmp] = -1;             }         }         return ans;     } }ac;   int main() {     cin.sync_with_stdio(0);     int T;     cin >> T;     while (T--)     {         ac.clear();         int n;         cin >> n;         for (int i = 0;i < n;i++)         {             string s;             cin >> s;             ac.insert(s);         }         ac.build();         string t;         cin >> t;         cout << ac.solve(t) << endl;     }     return 0; }