BZOJ 3277 串 题解

Statement

给 $n$ 个串，问每个串有多少子串是所有 $n$ 个串中至少 $k$ 个串的子串。

Solution 1

对于每个后缀，二分他最右的满足条件的前缀，这就是他贡献多少次。

设二分的 $x$，可以二分找出 height 上左边、右边第一个 $<$ 他的位置

同样预处理对于所有 $l$，最左的 $r$ 满足 $[l..r]$ 中出现了至少 $k$ 个串的后缀。

找到管辖区间 $[l..r]$ 后就可以 $O(1)$ 判了。

$O(n\log^2 n)$

Solution 2

考虑优化

考虑到原串中若后缀 $i-1$ 有 $x$ 个前缀是至少 $k$ 个串的子串，那么后缀 $i$ 有至少 $x-1$ 个前缀是至少 $k$ 个串的子串

按这样的方式算，是 $O(n\log n)$

Solution 3

建广义 SAM

设 $p(u)$ 为 $u$ 来自哪个串，$u$ 的出现次数就是他的子树内 $p$ 的种类数，这就是 HH 的项链

然后对于每个串的答案，在 SAM 上走一遍，每次若出现次数 $<k$ 就一直跳 $\text{link}$，直到出现次数 $\ge k$ 然后给答案加上 $\text{len}(u)$.

$O(n\log n)$

Code 1

log^2

// log^2
#include <bits/stdc++.h>
using namespace std;
#define rep(i, j, k) for (int i = (j); i <= (k); ++i)
#define reo(i, j, k) for (int i = (j); i >= (k); --i)
typedef long long ll;
const int N = 2e5 + 10, LN = 20;
string T, s[N];
int n, m = 27, p, t, k, sa[N], rk[N], se[N], cnt[N], height[N], belongs[N], suffixlength[N];
 
void Rsort() {
	rep(i, 1, m) cnt[i] = 0;
	rep(i, 1, n) ++cnt[rk[i]];
	rep(i, 1, m) cnt[i] += cnt[i - 1];
	reo(i, n, 1) sa[cnt[rk[se[i]]]--] = se[i];
}
void SA() {
	if (n == 1) return (void)(sa[1] = rk[1] = 1);
	rep(i, 1, n) rk[i] = ('a' <= T[i - 1] && T[i - 1] <= 'z') ? (T[i - 1] - 'a' + 1) : 27, se[i] = i;
	Rsort();
	for (int w = 1; w < n; w <<= 1, m = p) {
		p = 0;
		rep(i, n - w + 1, n) se[++p] = i;
		rep(i, 1, n) if (sa[i] > w) se[++p] = sa[i] - w;
		Rsort(), swap(rk, se), p = 0;
		rep(i, 1, n)
			if (se[sa[i]] == se[sa[i - 1]] && se[sa[i] + w] == se[sa[i - 1] + w]) rk[sa[i]] = p;
			else rk[sa[i]] = ++p;
		if (p == n) break;
	}
}
void Getheight() {
	int h = 0;
	rep(i, 1, n) {
		if (h) --h;
		while (T[i + h - 1] == T[sa[rk[i] - 1] + h - 1]) ++h;
		height[rk[i]] = h;
	}
}
int ln, Mn[N][LN];
void initST() {
	ln = __lg(n);
	rep(i, 1, n) Mn[i][0] = height[i];
	rep(j, 1, ln)
		rep(i, 1, n - (1 << j) + 1) Mn[i][j] = min(Mn[i][j - 1], Mn[i + (1 << (j - 1))][j - 1]);
}
int AskMn(int l, int r) {
	int g = __lg(r - l + 1);
	return min(Mn[l][g], Mn[r - (1 << g) + 1][g]);
}
 
ll ans[N];
int f[N];
void initF() {
	int r = 0;
	vector<int> buc(t + 1, 0);
	buc[0] = 114514;
	int res = 0;
	rep(l, 1, n) {
		while (r < n && res < k) {
			if (!buc[belongs[sa[++r]]]++) ++res;
		}
		if (res >= k) f[l] = r;
		else f[l] = 2e9;
		if (!--buc[belongs[sa[l]]]) --res;
	}
}
void Solve() {
	rep(i, 1, n) {
		int len = suffixlength[sa[i]];
		auto check = [&](int x) {
			int L = 1, R = n;
			{
				int l = 2, r = i, mid = 0, res = 1;
				while (l <= r) {
					mid = (l + r) >> 1;
					if (AskMn(mid, i) < x) res = mid, l = mid + 1;
					else r = mid - 1;
				}
				L = res;
			}
			{
				int l = i + 1, r = n, mid = 0, res = n + 1;
				while (l <= r) {
					mid = (l + r) >> 1;
					if (AskMn(i + 1, mid) < x) res = mid, r = mid - 1;
					else l = mid + 1;
				}
				R = res - 1;
			}
			return R >= f[L];
		};
		int l = 1, r = len, mid = 0, res = 0;
		while (l <= r) {
			mid = (l + r) >> 1;
			if (check(mid)) res = mid, l = mid + 1;
			else r = mid - 1;
		}
		ans[belongs[sa[i]]] += res;
	}
}
 
int main() {
	ios::sync_with_stdio(false), cin.tie(nullptr);
	cin >> t >> k;
	rep(i, 1, t) {
		cin >> s[i], T += s[i];
		if (i != t) T += '#';
	}
	n = T.length(), SA(), Getheight(), initST();
	int pos = 1;
	rep(i, 1, t) {
		int len = s[i].length();
		rep(j, pos, pos + len - 1)
			belongs[j] = i, suffixlength[j] = pos + len - 1 - j + 1;
		pos = pos + len + 1;
	}
	initF(), Solve();
	rep(i, 1, t) cout << ans[i] << " \n"[i == t];
	return 0;
}

Code 2

log，后缀数组

// log
#include <bits/stdc++.h>
using namespace std;
#define rep(i, j, k) for (int i = (j); i <= (k); ++i)
#define reo(i, j, k) for (int i = (j); i >= (k); --i)
typedef long long ll;
const int N = 4e5 + 10, LN = 20;
string s[N], T;
int n, m = 27, p, t, k, sa[N], rk[N], se[N], cnt[N], height[N];
 
void Rsort() {
	rep(i, 1, m) cnt[i] = 0;
	rep(i, 1, n) ++cnt[rk[i]];
	rep(i, 1, m) cnt[i] += cnt[i - 1];
	reo(i, n, 1) sa[cnt[rk[se[i]]]--] = se[i];
}
void SA() {
	if (n == 1) return (void)(sa[1] = rk[1] = 1);
	rep(i, 1, n) rk[i] = ('a' <= T[i - 1] && T[i - 1] <= 'z') ? (T[i - 1] - 'a' + 1) : 27, se[i] = i;
	Rsort();
	for (int w = 1; w < n; w <<= 1, m = p) {
		p = 0;
		rep(i, n - w + 1, n) se[++p] = i;
		rep(i, 1, n) if (sa[i] > w) se[++p] = sa[i] - w;
		Rsort(), swap(rk, se), p = 0;
		rep(i, 1, n)
			if (se[sa[i]] == se[sa[i - 1]] && se[sa[i] + w] == se[sa[i - 1] + w]) rk[sa[i]] = p;
			else rk[sa[i]] = ++p;
		if (p == n) break;
	}
}
int ln, Mn[N][LN];
void Getheight() {
	int k = 0;
	rep(i, 1, n) {
		if (k) --k;
		while (T[i + k - 1] == T[sa[rk[i] - 1] + k - 1]) ++k;
		height[rk[i]] = k;
	}
}
void initST() {
	ln = __lg(n);
	rep(i, 1, n) Mn[i][0] = height[i];
	rep(j, 1, ln)
		rep(i, 1, n - (1 << j) + 1)
			Mn[i][j] = min(Mn[i][j - 1], Mn[i + (1 << (j - 1))][j - 1]);
}
int AskMn(int l, int r) {
	int k = __lg(r - l + 1);
	return min(Mn[l][k], Mn[r - (1 << k) + 1][k]);
}
int f[N], belongs[N];
void initF() {
	int r = 0;
	vector<int> buc(t + 1, 0);
	buc[0] = 114514;
	int res = 0;
	rep(l, 1, n) {
		while (r < n && res < k) {
			if (!buc[belongs[sa[++r]]]++) ++res;
		}
		if (res >= k) f[l] = r;
		else f[l] = 2e9;
		if (!--buc[belongs[sa[l]]]) --res;
	}
}
ll ans[N];
int Suffixlength[N];
 
void Solve() {
	int k = 0;
	rep(i, 1, n) {
		if (k) --k;
		int len = s[belongs[i]].length();
		auto check = [&](int x) {
			int L = 1, R = n, pos = rk[i];
			{
				int l = 2, r = pos, mid = 0, res = 1;
				while (l <= r) {
					mid = (l + r) >> 1;
					if (AskMn(mid, pos) < x) res = mid, l = mid + 1;
					else r = mid - 1;
				}
				L = res;
			}
			{
				int l = pos + 1, r = n, mid = 0, res = n + 1;
				while (l <= r) {
					mid = (l + r) >> 1;
					if (AskMn(pos + 1, mid) < x) res = mid, r = mid - 1;
					else l = mid + 1;
				}
				R = res - 1;
			}
			return R >= f[L];
		};
		while (k < n - i + 1 && check(k + 1)) ++k;
		ans[belongs[i]] += min(Suffixlength[i], k);
	}
}
 
int main() {
	ios::sync_with_stdio(false), cin.tie(nullptr);
	cin >> t >> k;
	rep(i, 1, t) {
		cin >> s[i];
		T += s[i];
		if (i != t) T += '#';
	}
	n = T.length(), SA(), Getheight(), initST();
	int pos = 1;
	rep(i, 1, t) {
		int len = s[i].length();
		rep(j, pos, pos + len - 1) 
			belongs[j] = i, Suffixlength[j] = pos + len - 1 - j + 1;
		pos = pos + len + 1;
	}
	initF(), Solve();
	rep(i, 1, t) cout << ans[i] << " \n"[i == t];
	return 0;
}