洛谷 P5840 / LOJ 3733 「COCI 2015.1」DIVLJAK

洛谷传送门

LOJ 传送门

思路

对 \(S_1,S_2,...,S_n\) 建出 AC 自动机并建出 \(\mathrm{fail}\) 树。对于每次新加入的 \(P\)，考虑计算它对答案的贡献。

考虑在 AC 自动机上匹配的过程。加入一个 \(P\)，设它在 AC 自动机上从根结点开始的链经过 \(p_1,p_2,...,p_k\ (k = |P|)\)，那么它包含的子串为 \(p_1,p_2,...,p_k\) 到根结点的链经过的所有结点的并。

可以将 \(p_1,p_2,...,p_k\) 先按 dfn 序排序后，对于 \(i \in [1,k]\)，将 \(p_i\) 到根结点的链上所有权值 \(+1\)；对于 \(i \in [1,k-1]\)，将 \(\operatorname{lca}(p_i,p_{i+1})\) 到根结点的链上所有权值 \(-1\)。查询时直接查询对应结束结点的权值即可。这是一个树链加、单点求值的问题，差分后转化为单点加、子树求和，可以树状数组维护。

时间复杂度为 \(O(n \log n + 26n)\)。

代码

code

/*

p_b_p_b txdy
AThousandMoon txdy
AThousandSuns txdy
hxy txdy

*/

#include <bits/stdc++.h>
#define pb push_back
#define fst first
#define scd second

using namespace std;
typedef long long ll;
typedef pair<ll, ll> pii;

const int maxn = 2000100;

int n, m, idx[maxn], head[maxn], len, pp[maxn];
int dfn[maxn], times, fa[maxn], sz[maxn], dp[maxn], son[maxn];
int top[maxn], c[maxn];
bool vis[maxn];
char s[maxn];

struct edge {
	int to, next;
} edges[maxn << 1];

void add_edge(int u, int v) {
	edges[++len].to = v;
	edges[len].next = head[u];
	head[u] = len;
}

bool cmp(int a, int b) {
	return dfn[a] < dfn[b];
}

struct AC {
	int ch[maxn][26], tot, fail[maxn];
	
	void insert(char *s, int id) {
		int p = 0;
		for (int i = 0; s[i]; ++i) {
			if (!ch[p][s[i] - 'a']) {
				ch[p][s[i] - 'a'] = ++tot;
			}
			p = ch[p][s[i] - 'a'];
		}
		idx[id] = p;
	}
	
	void build() {
		queue<int> q;
		for (int i = 0; i < 26; ++i) {
			if (ch[0][i]) {
				q.push(ch[0][i]);
			}
		}
		while (q.size()) {
			int u = q.front();
			q.pop();
			for (int i = 0; i < 26; ++i) {
				if (ch[u][i]) {
					fail[ch[u][i]] = ch[fail[u]][i];
					q.push(ch[u][i]);
				} else {
					ch[u][i] = ch[fail[u]][i];
				}
			}
		}
		for (int i = 1; i <= tot; ++i) {
			add_edge(fail[i], i);
		}
	}
} ac;

int dfs(int u, int f, int d) {
	fa[u] = f;
	dp[u] = d;
	sz[u] = 1;
	dfn[u] = ++times;
	int maxson = -1;
	for (int i = head[u]; i; i = edges[i].next) {
		int v = edges[i].to;
		if (v == f) {
			continue;
		}
		sz[u] += dfs(v, u, d + 1);
		if (sz[v] > maxson) {
			son[u] = v;
			maxson = sz[v];
		}
	}
	return sz[u];
}

void dfs2(int u, int tp) {
	top[u] = tp;
	vis[u] = 1;
	if (!son[u]) {
		return;
	}
	dfs2(son[u], tp);
	for (int i = head[u]; i; i = edges[i].next) {
		int v = edges[i].to;
		if (!vis[v]) {
			dfs2(v, v);
		}
	}
}

int querylca(int x, int y) {
	while (top[x] != top[y]) {
		if (dp[top[x]] < dp[top[y]]) {
			swap(x, y);
		}
		x = fa[top[x]];
	}
	if (dp[x] > dp[y]) {
		swap(x, y);
	}
	return x;
}

inline int lowbit(int x) {
	return x & (-x);
}

void update(int x, int d) {
	for (int i = x; i <= times; i += lowbit(i)) {
		c[i] += d;
	}
}

int query(int x) {
	int res = 0;
	for (int i = x; i; i -= lowbit(i)) {
		res += c[i];
	}
	return res;
}

void solve() {
	scanf("%d", &n);
	for (int i = 1; i <= n; ++i) {
		scanf("%s", s);
		ac.insert(s, i);
	}
	ac.build();
	dfs(0, -1, 1);
	dfs2(0, 0);
	/*
	for (int i = 0; i <= ac.tot; ++i) {
		printf("%d ", dfn[i]);
	}
	putchar('\n');
	*/
	scanf("%d", &m);
	while (m--) {
		int op, x;
		scanf("%d", &op);
		if (op == 1) {
			scanf("%s", s + 1);
			int p = 0, len = strlen(s + 1);
			for (int i = 1; i <= len; ++i) {
				p = ac.ch[p][s[i] - 'a'];
				pp[i] = p;
			}
			sort(pp + 1, pp + len + 1, cmp);
			for (int i = 1; i <= len; ++i) {
				update(dfn[pp[i]], 1);
			}
			for (int i = 1; i < len; ++i) {
				update(dfn[querylca(pp[i], pp[i + 1])], -1);
			}
		} else {
			scanf("%d", &x);
			printf("%d\n", query(dfn[idx[x]] + sz[idx[x]] - 1) - query(dfn[idx[x]] - 1));
		}
	}
}

int main() {
	int T = 1;
	// scanf("%d", &T);
	while (T--) {
		solve();
	}
	return 0;
}