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AC自动机 - 多模式串匹配问题的基本运用 + 模板题 --- HDU 2222

2024-07-25 02:01:51   218人阅读

Keywords Search

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 35655    Accepted Submission(s): 11496


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
 

 

Mean: 

 给你n个单词，再给你一篇文章，让你统计有多少个单词在文章中出现过。

analyse:

裸的AC自动机，模板题。

Time complexity：o(n)+o(ml)    n个模式串长度均不超过m,文本串长度为L

 

Source code：

 

// Memory   Time// 1347K     0MS// by : Snarl_jsb// 2014-09-29-20.14#include<algorithm>#include<cstdio>#include<cstring>#include<cstdlib>#include<iostream>#include<vector>#include<queue>#include<stack>#include<map>#include<string>#include<climits>#include<cmath>#define N 10010#define LL long longusing namespace std;namespace ac_auto{    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)      //      插入单词    {        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)        //      bfs建立fail指针    {        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)       //  匹配 + 统计    {        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;    }}using namespace ac_auto;int main(){    int T, n;    scanf("%d",&T);    while(T--)    {        head = tail = 0;    //  清零        root = new node();      //  申请新的root结点        scanf("%d",&n);        while(n--)        {            scanf("%s", str);            Insert(str);    //  插入单词        }        build_ac_automation(root);      //  建树        scanf("%s",str);        printf("%d\n", Query(root));        //      查找+统计    }    return 0;}

　　

AC自动机 - 多模式串匹配问题的基本运用 + 模板题 --- HDU 2222


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