CodeForces 1801G A task for substrings

洛谷传送门

CF 传送门

区间显然不好处理，考虑转化成前缀和后缀。

设 \(f'_i\) 为 \(T[1 : i]\) 的单词出现次数，\(f_i\) 为 \(f'_i\) 的前缀和，\(g_i\) 为 \(T[1 : i]\) 后缀最长的单词编号。都可以通过建 \(s_i\) 正串的 ACAM 预处理。

对于询问 \((l, r)\)，一个简单的想法是直接回答 \(f_r - f_{l - 1}\)。但是我们多算了左端点在 \(l\) 之前的串。

考虑找到最大的 \(p\) 使得 \(p - |s_{g_p}| + 1 \le l\)，这样右端点在 \([p + 1, r]\) 的答案就是 \(f_r - f_p\)。再建反串 ACAM 预处理出 \(h_{i, j}\) 表示 \(s_i\) 长度为 \(j\) 的后缀有多少个串，就可以回答右端点在 \([l, p]\) 的答案。

找 \(p\) 可以线段树上二分。

code

// Problem: G. A task for substrings
// Contest: Codeforces - Codeforces Round 857 (Div. 1)
// URL: https://codeforces.com/contest/1801/problem/G
// Memory Limit: 1024 MB
// Time Limit: 4000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define pb emplace_back
#define fst first
#define scd second
#define mkp make_pair
#define mems(a, x) memset((a), (x), sizeof(a))

using namespace std;
typedef long long ll;
typedef double db;
typedef unsigned long long ull;
typedef long double ldb;
typedef pair<ll, ll> pii;

const int maxn = 1000100, maxm = 5000100;

int n, m, len, g[maxm];
ll f[maxm];
char t[maxm];
string s[maxn], rs[maxn];
vector<ll> h[maxn];

struct AC {
	int ch[maxn][26], fail[maxn], ntot, ed[maxn], id[maxn];
	
	inline void insert(string s, int k) {
		int p = 0;
		for (char c : s) {
			if (!ch[p][c - 'a']) {
				ch[p][c - 'a'] = ++ntot;
			}
			p = ch[p][c - 'a'];
		}
		ed[p] = 1;
		id[p] = k;
	}
	
	inline 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();
			ed[u] += ed[fail[u]];
			if (!id[u]) {
				id[u] = id[fail[u]];
			}
			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];
				}
			}
		}
	}
} ac, ca;

namespace SGT {
	int mn[maxm << 2];
	
	inline void pushup(int x) {
		mn[x] = min(mn[x << 1], mn[x << 1 | 1]);
	}
	
	void build(int rt, int l, int r) {
		if (l == r) {
			mn[rt] = l - (int)s[g[l]].size() + 1;
			return;
		}
		int mid = (l + r) >> 1;
		build(rt << 1, l, mid);
		build(rt << 1 | 1, mid + 1, r);
		pushup(rt);
	}
	
	int findr(int rt, int l, int r, int x) {
		if (mn[rt] > x) {
			return -1;
		}
		if (l == r) {
			return l;
		}
		int mid = (l + r) >> 1;
		return (mn[rt << 1 | 1] <= x) ? findr(rt << 1 | 1, mid + 1, r, x) : findr(rt << 1, l, mid, x);
	}
	
	int findr(int rt, int l, int r, int ql, int qr, int x) {
		if (mn[rt] > x) {
			return -1;
		}
		if (l == r) {
			return l;
		}
		int mid = (l + r) >> 1;
		if (qr > mid) {
			int t = findr(rt << 1 | 1, mid + 1, r, ql, qr, x);
			if (t != -1) {
				return t;
			}
		}
		if (ql <= mid) {
			int t = findr(rt << 1, l, mid, ql, qr, x);
			if (t != -1) {
				return t;
			}
		}
		return -1;
	}
}

void solve() {
	cin >> n >> m >> (t + 1);
	len = strlen(t + 1);
	for (int i = 1; i <= n; ++i) {
		cin >> s[i];
		rs[i] = s[i];
		reverse(rs[i].begin(), rs[i].end());
		ac.insert(s[i], i);
		ca.insert(rs[i], i);
	}
	ac.build();
	ca.build();
	int p = 0;
	for (int i = 1; i <= len; ++i) {
		p = ac.ch[p][t[i] - 'a'];
		g[i] = ac.id[p];
		f[i] = f[i - 1] + ac.ed[p];
	}
	for (int i = 1; i <= n; ++i) {
		h[i].resize(s[i].size());
		int p = 0;
		for (int j = 0; j < (int)s[i].size(); ++j) {
			if (j) {
				h[i][j] = h[i][j - 1];
			}
			p = ca.ch[p][rs[i][j] - 'a'];
			h[i][j] += ca.ed[p];
		}
	}
	SGT::build(1, 1, len);
	while (m--) {
		int l, r;
		cin >> l >> r;
		int p = SGT::findr(1, 1, len, l, r, l);
		if (p == -1) {
			printf("%lld ", f[r] - f[l - 1]);
		} else {
			printf("%lld ", f[r] - f[p] + h[g[p]][p - l]);
		}
	}
}

int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	int T = 1;
	// scanf("%d", &T);
	while (T--) {
		solve();
	}
	return 0;
}