Luogu P2336 [SCOI2012]喵星球上的点名

题目链接 \(Click Here\)_

\(200\)行纯干货的代码，一发\(WA\)掉真的是让人窒息，幸好最后找到了锅在哪。（差点就要弃掉了\(QAQ\)）

【调出来的时候真的是要高兴到哭出来了\(TwT\)】

这个题有很多种写法，但是大多数都是强行水过去的，正解的话应该是\(SA\)或者\(SAM\)搞一些数据结构维护。我在这里选择的是线段树维护\(RMQ\)，第一问就变成了区间颜色个数，第二问把第一问的答案差分对应到每个颜色上。放\(SA+\)莫队的人好像很少，这里我给一发清新脱俗又臭又长的代码供大家参考。

说句实话，如果我会\(SAM\)的话绝对不会用这种毒瘤写法。太\(TM\)难调了。

#include <bits/stdc++.h>
using namespace std;

const int N = 600010;
const int INF = 0x7fffffff;

#define ls (p << 1)
#define rs (p << 1 | 1)

int n, m, l, len, s[N], id[N], height[N];
int sa[N], tp[N], rk[N], _rk[N], bin[N], qry[N], lenq[N];

void base_sort (int n, int m) {
	for (int i = 1; i <= m; ++i) bin[i] = 0;
	for (int i = 1; i <= n; ++i) bin[rk[tp[i]]]++;
	for (int i = 1; i <= m; ++i) bin[i] += bin[i - 1];
	for (int i = n; i >= 1; --i) sa[bin[rk[tp[i]]]--] = tp[i];
}

void suffix_sort (int n, int m) {
	for (int i = 1; i <= n; ++i) {
		rk[i] = s[i], tp[i] = i;
	}
	base_sort (n, m);
	for (int w = 1; w <= n; w <<= 1) {
		int cnt = 0;
		for (int i = n - w + 1; i <= n; ++i) {
			tp[++cnt] = i;
		}
		for (int i = 1; i <= n; ++i) {
			if (sa[i] > w) {
				tp[++cnt] = sa[i] - w;
			}
		}
		base_sort (n, m);
		memcpy (_rk, rk, sizeof (rk));
		rk[sa[1]] = cnt = 1;
		for (int i = 2; i <= n; ++i) {
			rk[sa[i]] = _rk[sa[i]] == _rk[sa[i - 1]] && _rk[sa[i] + w] == _rk[sa[i - 1] + w] ? cnt : ++cnt;
		}
		if (cnt == n) return;
		m = cnt;
	}
}

void get_height (int n) {
	int k = 0;
	for (int i = 1; i <= n; ++i) {
		if (k) --k;
		int j = sa[rk[i] - 1];
		while (s[i + k] == s[j + k]) ++k;
		height[rk[i]] = k;
	}
}

struct Segment_Tree {
	//维护height数组最小值的线段树
	int a[N << 2];

	void push_up (int p) {
		a[p] = min (a[ls], a[rs]);
	}

	void build (int l, int r, int p) {
		if (l == r) {
			a[p] = height[l];
			return;
		}
		int mid = (l + r) >> 1;
		build (l, mid, ls);
		build (mid + 1, r, rs);
		push_up (p);
	}

	int query (int l, int r, int nl, int nr, int p) {
		if (nl <= l && r <= nr) {
			return a[p];
		}
		int mid = (l + r) >> 1, res = INF;
		if (nl <= mid) res = min (res, query (l, mid, nl, nr, ls));
		if (mid + 1 <= nr)  res = min (res, query (mid + 1, r, nl, nr, rs));
		return res;
	}
}tr;

struct Query {
	int l, r, id, blo;
}q[N];

bool cmp (Query lhs, Query rhs) {
	return lhs.blo == rhs.blo ? lhs.r < rhs.r : lhs.blo < rhs.blo;
}

bool can_use (int l, int r, int lcp) {
	return tr.query (1, len, l, r, 1) >= lcp;
}

int have[N], ans1[N], ans2[N];

int main () {
	cin >> n >> m;
	int ban = 10001;
	s[len = 0] = -1;
	for (int i = 1; i <= n; ++i) {
		for (int k = 0; k <= 1; ++k) {
			cin >> l;
			for (int j = 1; j <= l; ++j) {
				cin >> s[++len];
				id[len] = i;
			}
			s[++len] = ++ban;
		}
	}
	for (int i = 1; i <= m; ++i) {
		cin >> l;
		lenq[i] = l;
		qry[i] = len + 1;
		for (int j = 1; j <= l; ++j) {
			cin >> s[++len];
		}
		s[++len] = ++ban;
	}
	suffix_sort (len, ban);
	get_height (len);
	tr.build (1, len, 1);
	int maxn = sqrt (len);
	for (int i = 1; i <= m; ++i) {
		q[i].id = i;
		int l = 1, r = rk[qry[i]];
		while (l < r) {
			int mid = (l + r) >> 1;
			if (can_use (mid, r, lenq[i])) {
				r = mid;
			} else {
				l = mid + 1;
			}
		}
		if (height[r] < lenq[i]) r++;
		q[i].l = r - 1;
		//确认左边界
		l = rk[qry[i]], r = len;
		while (l < r) {
			int mid = (l + r + 1) >> 1;
			if (can_use (l, mid, lenq[i])) {
				l = mid;
			} else {
				r = mid - 1;
			}
		}
		if (height[l] < lenq[i]) l--;
		q[i].r = l;
		//确认右边界
		q[i].blo = (q[i].l - 1) / maxn + 1;
		//分块
	}
	sort (q + 1, q + 1 + m, cmp);
	int l = 1, r = 0, tot = 0;
	for (int i = 1; i <= m; ++i) {
		while (l < q[i].l) {
 			--have[id[sa[l]]];
			if (id[sa[l]] != 0 && have[id[sa[l]]] == 0) {
				--tot;
				ans2[id[sa[l]]] -= (m - i + 1);
			}
			++l;
		}
		while (l > q[i].l) {
			--l;
			if (id[sa[l]] != 0 && have[id[sa[l]]] == 0) {
				++tot;
				ans2[id[sa[l]]] += (m - i + 1);
			}
			++have[id[sa[l]]];
		}
		while (r > q[i].r) {
			--have[id[sa[r]]];
			if (id[sa[r]] != 0 && have[id[sa[r]]] == 0) {
				--tot;
				ans2[id[sa[r]]] -= (m - i + 1);
			}
			--r;
		}
		while (r < q[i].r) {
			++r;
			if (id[sa[r]] != 0 && have[id[sa[r]]] == 0) {
				++tot;
				ans2[id[sa[r]]] += (m - i + 1);
			}
			++have[id[sa[r]]];
		}
		ans1[q[i].id] = tot;
	}
	for (int i = 1; i <= m; ++i) printf ("%d\n", ans1[i]);
	for (int i = 1; i <= n; ++i) printf ("%d ", ans2[i]);
}