HDU 2222 keywords search (AC automatic machine template)

Keywords search Time Limit: 2000/1000 MS (Java/others) memory limit: 65536/32768 K (Java/Others)

Problem descriptionin 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 shoshould tell me how many keywords will be match.
 
Inputfirst line will contain one integer means how many cases will follow.
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 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

15shehesayshrheryasherhs

 
Sample output

3

Question: give n words and a text string, and ask how many words have appeared in the text string. Analysis: there is no need for any conversion for the entry Question of the AC automatic machine. Good reference for learning AC automatic machine: http://www.cs.uku.fi /~ Kilpelai/bsa05/lectures/slides04.pdf

#include<cstdio>const int N = 10010;char str[1000005];struct node{    node *next[26];    node *fail;    int count;    node() {        for(int i = 0; i < 26; i++)            next[i] = NULL;        count = 0;        fail = NULL;    }}*q[50*N];node *root;int head, tail;void Insert(char *str){    node *p = root;    int i = 0, index;    while(str[i]) {        index = str[i] - 'a';        if(p->next[index] == NULL)            p->next[index] = new node();        p = p->next[index];        i++;    }    p->count++;}void build_ac_automation(node *root){    root->fail = NULL;    q[tail++] = root;    while(head < tail) {        node *temp = q[head++];        node *p = NULL;        for(int i = 0; i < 26; i++) {            if(temp->next[i] != NULL) {                if(temp == root) temp->next[i]->fail = root;                else {                    p = temp->fail;                    while(p != NULL) {                        if(p->next[i] != NULL) {                            temp->next[i]->fail = p->next[i];                            break;                        }                        p = p->fail;                    }                    if(p == NULL) temp->next[i]->fail = root;                }                q[tail++] = temp->next[i];            }        }    }}int Query(node *root){    int i = 0, cnt = 0, index;    node *p = root;    while(str[i]) {        index = str[i] - 'a';        while(p->next[index] == NULL && p != root) p = p->fail;        p = p->next[index];        if(p == NULL) p = root;        node *temp = p;        while(temp != root && temp->count != -1) {            cnt += temp->count;            temp->count = -1;            temp = temp->fail;        }        i++;    }    return cnt;}int main(){    int T, n;    scanf("%d",&T);    while(T--) {        head = tail = 0;        root = new node();        scanf("%d",&n);        while(n--) {            scanf("%s", str);            Insert(str);        }        build_ac_automation(root);        scanf("%s",str);        printf("%d\n", Query(root));    }    return 0;}