Codeforces 163E（ac自动机、树状数组）

要点

• 显然ac自动机的板子就可以暴力一下答案了
• 为了优化时间复杂度，考虑套路fail树的dfs序。发现本题需要当前这个尾点加上所有祖先点的个数，考虑使用树状数组差分一下，在父点+1，在子树后-1，每次询问前缀和即可

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
using namespace std;
typedef long long ll;
const int N = 1e6 + 5, maxk = 1e5 + 5;
int n, k, pos[maxk], in[maxk];
string s, t;
struct FenwickTree {
	ll F[N];
	void Modify(int x, int val) {
		for (; x < N; x += x&-x)
			F[x] += val;
	}
	ll Query(int x) {
		ll res = 0;
		for (; x; x -= x&-x)
			res += F[x];
		return res;
	}
}bit;
struct Aho_Corasick_Automaton {
    int ch[N][26];//Trie树的转移
    //int val[N];//根据题意赋值。有值则意味着某子串末尾
    int fail[N];//失配，转移到别的树枝接着找
    int sz;//注意这个板子sz一定是要从1开计
    vector<int> adj[N];//用于fail树
    int dfn[N], Time, size[N];
    void insert(string s, int id) {//Trie的插入
        int len = s.length(), now = 0;
        for (int i = 0; i < len; i++){
            int c = s[i] - 'a';
            if (!ch[now][c]) {
                sz++;
                memset(ch[sz], 0, sizeof ch[sz]);
                //val[sz] = 0;
                ch[now][c] = sz;
            }
            now = ch[now][c];
        }
        //val[now] = id;
        pos[id] = now;
    }
    void getfail(){
        queue<int> Q;
        for (int i = 0; i < 26; i++)
            if (ch[0][i]) {
            	adj[0].emplace_back(ch[0][i]);
                fail[ch[0][i]] = 0, Q.push(ch[0][i]);//第二层指向根
            }
        while (!Q.empty()){
            int u = Q.front(); Q.pop();
            for (int i = 0; i < 26; i++)
                if (ch[u][i]) {
                	adj[ch[fail[u]][i]].emplace_back(ch[u][i]);
                    fail[ch[u][i]] = ch[fail[u]][i], Q.push(ch[u][i]);//指向其他枝上同样的字母
                }
                else ch[u][i] = ch[fail[u]][i];//使得find时半路突然失配时还能一下拐回去
        }
    }
    void dfs(int u) {
    	dfn[u] = ++Time;
    	size[u] = 1;
    	for (int v : adj[u]) {
    		dfs(v);
    		size[u] += size[v];
    	}
    }
    int find(string T){
        int len = T.length(), now = 0;
        ll res = 0;
        for (int i = 0; i < len; i++){
            now = ch[now][T[i] - 'a'];
            res += bit.Query(dfn[now]);
            // for (int t = now; t; t = fail[t])//事实上这是一个不断缩短后缀的过程
            //     if (val[t]) res++;
        }
        return res;
    }
}ac;
int calc(string t) {
	stringstream ss(t);
	int x; ss >> x;
	return x;
}
int main() {
	ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	cin >> n >> k;
	for (int i = 1; i <= k; i++) {
		cin >> s;
		ac.insert(s, i);
		in[i] = 1;
	}
	ac.getfail();
	ac.dfs(0);
	for (int i = 1; i <= k; i++) {
		int id = pos[i];
		bit.Modify(ac.dfn[id], 1);
		bit.Modify(ac.dfn[id] + ac.size[id], -1);
	}
	for (int i = 0; i < n; i++) {
		cin >> t;
		char x = t[0]; t.erase(0, 1);
		if (x == '+') {
			int v = calc(t);
			if (!in[v]) {
				int id = pos[v];
				bit.Modify(ac.dfn[id], 1);
				bit.Modify(ac.dfn[id] + ac.size[id], -1);
				in[v] = 1;
			}
		} else if (x == '-') {
			int v = calc(t);
			if (in[v]) {
				int id = pos[v];
				bit.Modify(ac.dfn[id], -1);
				bit.Modify(ac.dfn[id] + ac.size[id], 1);
				in[v] = 0;
			}
		} else {
			cout << ac.find(t) << '\n';
		}
	}
}