P4093 序列 題解

Statement

給出 \(n\) 個數的序列 \(\{a_i\}\)，接下來 \(m\) 秒中每一秒會有一個數發生變化，然後恢復。

問最長的子序列長度，使得任意時刻這個子序列不下降。\(n\le 10^5\)

Solution

設 \(b_i\) 為 \(i\) 最小能變成的數，\(c_i\) 為 \(i\) 最大能變成的數

\[ f(i)=\max_{j<i\land c_j\le a_i\land a_j\le b_i}\{f(j)\}+1 \]

分治即可，\(O(n\log^2 n)\)

Code

#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;
int n, m, ans;
struct Item {
	int a, b, c, f, id;
} a[N];

struct BIT {
	int mx[N];
	void Upd(int x, int v) {
		for (; x <= 100000; x += x & -x) mx[x] = max(mx[x], v);
	}
	int Qry(int x) {
		int res = 0;
		for (; x; x -= x & -x) res = max(res, mx[x]);
		return res;
	}
	void Clear(int x) {
		for (; x <= 100000; x += x & -x) mx[x] = 0;
	}
} bit;

void Solve(int l, int r) {
	if (l == r) return (void)(ans = max(ans, a[l].f));
	int mid = (l + r) >> 1;
	Solve(l, mid);
	sort(a + l, a + mid + 1, [&](Item l, Item r) { return l.c < r.c; });
	sort(a + mid + 1, a + r + 1, [&](Item l, Item r) { return l.a < r.a; });
	for (int j = l, i = mid + 1; j <= mid || i <= r; ) {
		if (j <= mid && (a[j].c <= a[i].a || i > r)) {
			bit.Upd(a[j].a, a[j].f), ++j;
		} else {
			a[i].f = max(a[i].f, bit.Qry(a[i].b) + 1), ++i;
		}
	}
	rep(i, l, mid) bit.Clear(a[i].a);
	sort(a + l, a + r + 1, [&](Item l, Item r) { return l.id < r.id; });
	Solve(mid + 1, r);
}

int main() {
	ios::sync_with_stdio(false), cin.tie(nullptr);
	cin >> n >> m;
	rep(i, 1, n) {
		cin >> a[i].a, a[i].b = a[i].c = a[i].a, a[i].f = 1, a[i].id = i;
	}
	rep(i, 1, m) {
		int x, y;
		cin >> x >> y;
		a[x].b = min(a[x].b, y), a[x].c = max(a[x].c, y);
	}
	Solve(1, n);
	cout << ans << '\n';
	return 0;
}