hdu_2222 Keywords Search（AC自动机）

Keywords Search

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 58586    Accepted Submission(s): 19232

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
Aho-Corasick自动机模板题。
用cot数组保存单词出现的次数。
#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <stack>#include <bitset>#include <queue>#include <set>#include <map>#include <string>#include <algorithm>#define FOP freopen("data.txt","r",stdin)#define FOP2 freopen("data1.txt","w",stdout)#define inf 0x3f3f3f3f#define maxn 1000010#define mod 1000000007#define PI acos(-1.0)#define LL long longusing namespace std;struct ACAutomata{    int next[maxn][26], fail[maxn];    int cot[maxn];    int root, L;    int newnode()    {        for(int i = 0; i < 26; i++) next[L][i] = -1;        cot[L++] = 0;        return L-1;    }    void init()    {        L = 0;        root = newnode();    }    void insert(char buf[])    {        int len = strlen(buf);        int now = root;        for(int i = 0; i < len; i++)        {            if(next[now][buf[i]-'a'] == -1)                next[now][buf[i]-'a'] = newnode();            now = next[now][buf[i]-'a'];        }        cot[now]++;    }    void build()    {        queue<int> Q;        fail[root] = root;        for(int i = 0; i < 26; i++)        {            if(next[root][i] == -1)                next[root][i] = root;            else            {                fail[next[root][i]] = root;                Q.push(next[root][i]);            }        }        while(!Q.empty())        {            int now = Q.front();            Q.pop();            for(int i = 0; i < 26; i++)            {                if(next[now][i] == -1)                    next[now][i] = next[fail[now]][i];                else                {                    fail[next[now][i]] = next[fail[now]][i];                    Q.push(next[now][i]);                }            }        }    }    int query(char buf[])    {        int res = 0;        int len = strlen(buf);        int now = root;        for(int i = 0; i < len; i++)        {            now = next[now][buf[i]-'a'];            int temp = now;            while(temp != root)            {                res += cot[temp];                cot[temp] = 0;                temp = fail[temp];            }        }        return res;    }};int n;ACAutomata aca;char buf[maxn];int main(){    int T;    scanf("%d", &T);    while(T--)    {        aca.init();        scanf("%d", &n);        char s[52];        for(int i = 0; i < n; i++)        {            scanf("%s", s);            aca.insert(s);        }        aca.build();        scanf("%s", buf);        printf("%d\n", aca.query(buf));    }    return 0;}