【AC自动机】HDU-2222 Keywords Search

Keywords Search

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

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

15shehesayshrheryasherhs

 

Sample Output

3

 

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思路：AC自动机的模板题，用来测试模板。

P.S. 本题很坑的地方是这两组数据：

2

6
she
he
he
say
shr
her
yasherhs

Output: 4
6
she
he
he
say
shr
her
yasherhe

Output: 4

代码如下：

#include <cstdio>#include <cstdlib>#include <cstring>#include <queue>using namespace std;const int N = 5e5+5, M = 1e6+5;struct Trie {    Trie *next[26];    Trie *fail;    int val;}tree[N];queue <Trie *> q;int idx[128];char txt[M];class ac_auto {    private:        int nxt;        Trie *root;    public:        ac_auto() {            nxt = 0;            root = add();        }        Trie *add() {            memset(&tree[nxt], 0, sizeof(Trie));            return &tree[nxt++];        }        void insert(char *s) {            Trie *rt = root;            int len = strlen(s);            for(int i = 0; i < len; i++) {                int c = idx[s[i]+0];                if(!rt->next[c])                    rt->next[c] = add();                rt = rt->next[c];            }            rt->val++;        }        void get_f() {            queue <Trie *> q;            q.push(root);            root->fail = NULL;            while(!q.empty()) {                Trie *u = q.front(); q.pop();                for(int c = 0; c < 26; c++) {                    if(u->next[c]) {                        Trie *f = u->fail;                        while(f) {                            if(f->next[c]) {                                u->next[c]->fail = f->next[c];                                break;                            }                            f = f->fail;                        }                        if(!f) u->next[c]->fail = root;                        q.push(u->next[c]);                    }                }            }        }        int match(char *s) {            Trie *rt = root;            int len = strlen(s), ret = 0;            for(int i = 0; i < len; i++) {                int c = idx[s[i]+0];                while(!rt->next[c] && rt != root) rt = rt->fail;                rt = rt->next[c];                if(!rt) rt = root;                Trie *p = rt;                while(p != root) {                    if(p->val) {                        ret += p->val;                        p->val = 0;                    }                    else break;                    p = p->fail;                }            }            return ret;        }};void haxi(){    for(int i = 0; i < 26; i++)         idx['a'+i] = i;}int main(){#ifdef J_Sure    freopen("000.in", "r", stdin);    //freopen("999.out", "w", stdout);#endif    int T;    scanf("%d", &T);    haxi();    while(T--) {        ac_auto ac;        int m;        char str[55];        scanf("%d", &m);        while(m--) {            scanf("%s", str);            ac.insert(str);        }        ac.get_f();        scanf("%s", txt);        int ans = ac.match(txt);        printf("%d\n", ans);    }       return 0;}