Competitive Programming

/**
 * Author: Dean
 * License: CC0
 * Description: Builds an Ahocorasick Trie, with suffix links.
 * Time: O(n + m + z)
 * Status: Tested http://www.spoj.com/problems/STRMATCH/, https://codeforces.com/gym/101741/problem/K
 * Usage: This is an offline Algorithm, Pass the vector of patterns to Trie.init(v), find function returns the number of times each string appears
 */

const int MaxM = 200005;

struct Trie{
    static const int Alpha = 26;
    static const int first = 'a';
    int lst = 1;
    struct node{
        int nxt[Alpha] = {}, p = -1;
        char c;
        vector<int> end; //if all patterns are different, can use flag instead
        int SuffixLink = -1;
        int cnt = 0;
	};
	vector<node> V;
	int num;
	stack<int> reversebfs;
	inline int getval(char c) {
		return c - first;
	}
	void CreateSuffixLink() {
		queue<int> q;
		for(q.push(0); q.size(); q.pop()) {
			int pos = q.front();
			reversebfs.push(pos);
      if(!pos || !V[pos].p) V[pos].SuffixLink = 0;
      else {
        int val = getval(V[pos].c);
        int j = V[V[pos].p].SuffixLink;
        V[pos].SuffixLink = V[j].nxt[val];
      }
			for(int i = 0; i < Alpha; ++i) {
				if(V[pos].nxt[i]) q.push(V[pos].nxt[i]);
				else if(!pos || !V[pos].p) 	V[pos].nxt[i] = V[0].nxt[i];
				else V[pos].nxt[i] = V[V[pos].SuffixLink].nxt[i];
			}
		}
	}

	void init(vector<string> &v) {
		V.resize(MaxM);
		num = v.size();
		int id = 0;
		for(auto &s : v) {
			int pos = 0;
			for(char &c : s) {
				int val = getval(c);
				if(!V[pos].nxt[val]) {
					V[lst].p = pos; V[lst].c = c; V[pos].nxt[val] = lst++;
				}
				pos = V[pos].nxt[val];
			}
			V[pos].end.emplace_back(id++);
		}
		CreateSuffixLink();
	}

	vector<int> find(string& word) {
		int pos = 0;
		vector<int> ans(num, 0);
		for(auto &c : word) {
			int val = getval(c);
			pos = V[pos].nxt[val];
			V[pos].cnt++; //We count the times we reach each node, and then do a reverse propagation
		}
		for(;reversebfs.size();reversebfs.pop()) {
			int	x = reversebfs.top(); //When we process x, we know we have been there V[x].cnt times;
			for(int i : V[x].end) ans[i] += V[x].cnt;
			if(V[x].SuffixLink != -1) V[V[x].SuffixLink].cnt += V[x].cnt;
		}
		return ans;
	}
};