[CF 666E] Forensic Examination

Description

传送门

Solution

对 \(T[1..m]\) 建立广义后缀自动机，离线，找出代表 \(S[pl,pr]\) 的每个节点，线段树合并。

Code

#include <cstdio>
#include <vector>
#include <cstring>
#include <algorithm>

const int N = 500005, M = 9500005;
struct Edge {
	int v, nxt;
} e[N];
struct Node {
	int l, r, x, y, id;
	bool operator < (const Node & rhs) const {
		return y < rhs.y;
	}
} a[N];
struct Pair {
	int x, y;
	bool operator < (const Pair & rhs) const {
		return (x == rhs.x && y > rhs.y) || x < rhs.x;
	}
} sum[M], ans[N];
int L[N], q, n, m, mx[N], ch[N][30], sz = 1, las = 1, head[N], tot, fa[N], cnt, ls[M], rs[M], f[20][N], rt[N];
char s[N], t[N];
std::vector<int> b[N];

int read() {
	int x = 0; char c = getchar();
	while (c < '0' || c > '9') c = getchar();
	while (c >= '0' && c <= '9') x = (x << 3) + (x << 1) + (c ^ 48), c = getchar();
	return x;
}
void adde(int u, int v) {
	e[++tot].nxt = head[u], head[u] = tot, e[tot].v = v;
}
void ins(int c) {
	int p = las, np = ++sz;
	mx[np] = mx[p] + 1, las = np;
	for (; p && !ch[p][c]; p = fa[p]) ch[p][c] = np;
	if (!p) { fa[np] = 1; return; }
	int q = ch[p][c];
	if (mx[p] + 1 == mx[q]) { fa[np] = q; return; }
	int nq = ++sz;
	mx[nq] = mx[p] + 1, fa[nq] = fa[q], fa[np] = fa[q] = nq;
	memcpy(ch[nq], ch[q], sizeof ch[q]);
	for (; p && ch[p][c] == q; p = fa[p]) ch[p][c] = nq;
}
void update(int &x, int l, int r, int p) {
	if (!x) x = ++cnt;
	if (l == r) { sum[x].x = 1, sum[x].y = l; return; }
	int mid = (l + r) >> 1;
	if (p <= mid) update(ls[x], l, mid, p);
	else update(rs[x], mid + 1, r, p);
	sum[x] = std::max(sum[ls[x]], sum[rs[x]]);
}
Pair query(int x, int l, int r, int L, int R) {
	Pair res; res.x = 0, res.y = 0;
	if (!x) return res;
	if (L <= l && r <= R) return sum[x];
	int mid = (l + r) >> 1;
	if (L <= mid) res = query(ls[x], l, mid, L, R);
	if (mid < R) res = std::max(res, query(rs[x], mid + 1, r, L, R));
	return res;
}
int merge(int x, int y) {
	if (!x) return y;
	if (!y) return x;
	if (!ls[x] && !rs[x] && !ls[y] && !rs[y]) { sum[x].x += sum[y].x; return x; }
	ls[x] = merge(ls[x], ls[y]);
	rs[x] = merge(rs[x], rs[y]);
	sum[x] = std::max(sum[ls[x]], sum[rs[x]]);
	return x;
}
void dfs1(int u) {
	f[0][u] = fa[u];
	for (int i = 1; i <= 18; ++i) f[i][u] = f[i - 1][f[i - 1][u]];
	for (int i = head[u]; i; i = e[i].nxt) dfs1(e[i].v);
}
void jump(int x, int k) {
	for (int i = 18; ~i; --i)
		if (mx[f[i][x]] >= a[k].y - a[k].x + 1) x = f[i][x];
	b[x].push_back(k);
}
void dfs2() {
	int x = 1, y = 0;
	for (int i = 1, j = 1; i <= n; ++i) {
		int c = s[i - 1] - 'a';
		while (x && !ch[x][c]) x = fa[x], y = mx[x];
		if (!x) { x = 1, y = 0; continue; }
		++y, x = ch[x][c];
		while (j <= q && a[j].y < i) ++j;
		while (j <= q && a[j].y == i) {
			if (y >= a[j].y - a[j].x + 1) jump(x, j);
			++j;
		}
	}
}
void dfs3(int u) {
	for (int i = head[u]; i; i = e[i].nxt)
		dfs3(e[i].v), rt[u] = merge(rt[u], rt[e[i].v]);
	int sz = b[u].size();
	for (int i = 0; i < sz; ++i) ans[a[b[u][i]].id] = query(rt[u], 1, m, a[b[u][i]].l, a[b[u][i]].r);
}
int main() {
	scanf("%s", s), m = read();
	for (int i = 1; i <= m; ++i) {
		scanf("%s", t), n = strlen(t), las = 1;
		for (int j = 0; j < n; ++j)
			ins(t[j] - 'a'), update(rt[las], 1, m, i);
	}
	q = read(), n = strlen(s);
	for (int i = 1; i <= q; ++i) L[i] = a[i].l = read(), a[i].r = read(), a[i].x = read(), a[i].y = read(), a[i].id = i;
	std::sort(a + 1, a + q + 1);
	for (int i = 2; i <= sz; ++i) adde(fa[i], i);
	dfs1(1), dfs2(), dfs3(1);
	for (int i = 1; i <= q; ++i) printf("%d %d\n", ans[i].y ? ans[i].y : L[i], ans[i].x);
	return 0;
}