首页 > 代码库 > hdu 2222 Keywords Search（AC自动机模板题）

hdu 2222 Keywords Search（AC自动机模板题）

2024-07-17 18:50:59   221人阅读

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
1
5
she
he
say
shr
her
yasherhs
 

Sample Output
3
 
题意：给出n个单词和一个文本串，问在文本串中有多少个单词出现过。
分析：AC自动机入门题，不需要任何转化。
学习AC自动机较好的参考资料：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;
}


联系
我们