HDU 2222 Keywords Search AC自动机

Posted 2020-10-04 Hyouka

Keywords Search

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 67122    Accepted Submission(s): 22584

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

 

 

Author

Wiskey

 

 

最裸的AC自动机

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <vector>
#include <iomanip>
#include <math.h>
#include <map>
using namespace std;
#define FIN     freopen("input.txt","r",stdin);
#define FOUT    freopen("output.txt","w",stdout);
#define INFLL   0x3f3f3f3f3f3f3f
#define lson    l,m,rt<<1
#define rson    m+1,r,rt<<1|1
typedef long long LL;
typedef pair<double, double> PII;

struct Trie {
    int next[500010][26], fail[500010], end[500010];
    int root, L;
    int newnode() {
        for(int i = 0; i < 26; i++)
            next[L][i] = -1;
        end[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‘];
        }
        end[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 len = strlen(buf);
        int now = root;
        int res = 0;
        for(int i = 0; i < len; i++) {
            now = next[now][buf[i] - ‘a‘];
            int temp = now;
            while( temp != root ) {
                res += end[temp];
                end[temp] = 0;
                temp = fail[temp];
            }
        }
        return res;
    }
    void debug() {
        for(int i = 0; i < L; i++) {
            printf("id = %3d,fail = %3d,end = %3d,chi = [", i, fail[i], end[i]);
            for(int j = 0; j < 26; j++)
                printf("%2d", next[i][j]);
            printf("]\n");
        }
    }
};

Trie AC;
char s[1000005];

int main() {
    //FIN
    int T;
    scanf("%d", &T);
    while(T--) {
        AC.init();
        int n;
        scanf("%d", &n);
        for(int i = 1; i <= n; i++) {
            scanf("%s", s);
            AC.insert(s);
        }
        AC.build();
        scanf("%s", s);
        printf("%d\n", AC.query(s));
    }


    return 0;
}