HDU - 2222 Keywords Search AC自动机

  

Keywords Search

 HDU - 2222

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

Source

HDU - 2222

My Solution

题意：给出n个字符串为这些字符串在主串s中出现的个数。

AC自动机

裸的AC自动机，注意下给定模式串可能有一些相同的串，然后按照主串在自动机上遍历即可。

复杂度 O(n)

#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <queue>#include <map>using namespace std;typedef long long LL;const int CHAR_SIZE = 26;const int MAX_SIZE = 5e5 + 8;map<char, int> mp;struct AC_Machine{    int ch[MAX_SIZE][CHAR_SIZE], danger[MAX_SIZE], fail[MAX_SIZE];    int sz;    inline void init(){        sz = 1;        memset(ch[0], 0, sizeof ch[0]);        memset(danger, 0, sizeof danger);    }    inline void _insert(const string &s){        int n = s.size();        int u = 0, c;        for(int i = 0; i < n; i++){            c = mp[s[i]];            if(!ch[u][c]){                memset(ch[sz], 0, sizeof ch[sz]);                danger[sz] = 0;                ch[u][c] = sz++;            }            u = ch[u][c];        }        danger[u]++;    }    inline void _build(){        queue<int> Q;        fail[0] = 0;        for(int c = 0, u; c < CHAR_SIZE; c++){            u = ch[0][c];            if(u){Q.push(u); fail[u] = 0;}        }        int r;        while(!Q.empty()){            r = Q.front();            Q.pop();            //danger[r] |= danger[fail[r]];            for(int c = 0, u; c < CHAR_SIZE; c++){                u = ch[r][c];                if(!u){ch[r][c] = ch[fail[r]][c]; continue; }                fail[u] = ch[fail[r]][c];                Q.push(u);            }        }    }}ac;string s;int main(){    #ifdef LOCAL    freopen("3.in", "r", stdin);    //freopen("3.out", "w", stdout);    #endif // LOCAL    ios::sync_with_stdio(false); cin.tie(0);    int T, n, len, now, i, ans, tmp;    for(int i = 0; i < 26; i++){mp[(i + 'a')] = i;}    cin >> T;    while(T--){        cin >> n;        ac.init();        while(n--){            cin >> s;            ac._insert(s);        }        ac._build();        cin >> s;        ans = 0; len = s.size(); now = 0;        for(i = 0; i < len; i++){            now = ac.ch[now][mp[s[i]]];            tmp = now;            while(tmp){                if(ac.danger[tmp]){                    ans += ac.danger[tmp];                    ac.danger[tmp] = 0;                }                tmp = ac.fail[tmp];            }        }        cout << ans << "\n";    }    return 0;}

Thank you!

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