[洛谷P3966][TJOI2013]单词

题目大意：有$n$个字符串，求每个字符串在所有字符串中出现的次数

题解：$AC$自动机，每个节点被经过时$sz$加一，每一个字符串出现次数为其$fail$树子树$sz$和

卡点：$AC$自动机根节点为$1$，没有在$build$的时候将所有空的$nxt[1][i]$赋值为$1$

C++ Code：

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <queue>
const int maxn = 1e6 + 10;

std::string s;
int n, ret[210];
namespace AC {
	int nxt[maxn][26], fail[maxn], idx = 1, cnt[maxn];
	int insert(std::string s) {
		int p = 1;
		for (char ch : s) {
			if (nxt[p][ch - 'a']) p = nxt[p][ch - 'a'];
			else p = nxt[p][ch - 'a'] = ++idx;
			++cnt[p];
		}
		return p;
	}
	void build() {
		static std::queue<int> q;
		for (int i = 0; i < 26; ++i)
			if (nxt[1][i]) fail[nxt[1][i]] = 1, q.push(nxt[1][i]);
			else nxt[1][i] = 1;
		while (!q.empty()) {
			int u = q.front(); q.pop();
			for (int i = 0; i < 26; ++i)
				if (nxt[u][i]) fail[nxt[u][i]] = nxt[fail[u]][i], q.push(nxt[u][i]);
				else nxt[u][i] = nxt[fail[u]][i];
		}
	}

	int head[maxn], CNT;
	struct Edge {
		int to, nxt;
	} e[maxn];
	void addedge(int a, int b) {
		e[++CNT] = (Edge) { b, head[a] }; head[a] = CNT;
	}
	void dfs(int u) {
		for (int i = head[u], v; i; i = e[i].nxt) {
			v = e[i].to;
			dfs(v);
			cnt[u] += cnt[v];
		}
	}
	void solve() {
		for (int i = 2; i <= idx; ++i) addedge(fail[i], i);
		dfs(1);
	}
}
int main() {
	std::ios::sync_with_stdio(false), std::cin.tie(0), std::cout.tie(0);
	std::cin >> n;
	for (int i = 0; i < n; ++i) {
		std::cin >> s;
		ret[i] = AC::insert(s);
	}
	AC::build(), AC::solve();
	for (int i = 0; i < n; ++i) std::cout << AC::cnt[ret[i]] << '
';
	return 0;
}