UOJ#410. 【IOI2018】会议

传送门

首先可以设 \(f[l][r]\) 表示 \([l,r]\) 的答案

设 \(x\) 为区间 \([l,r]\) 的最大值的位置，那么

\(f[l][r] = min(f[l][x-1]+h[x]\times (r-x+1),f[x+1][r]+h[x]\times (x-l+1))\)

这样的 \(dp\) 结构形成了笛卡尔树

那么考虑在笛卡尔树上维护 \(dp\)

对于笛卡尔树上的一个点 \(x\) 它有一个区间 \([l,r]\)，考虑维护 \(f[l][i]\) 和 \(f[i][r],i\in [l,r]\)

对于固定左端点的 \(f[l][i]\)(\(f[i][r]\) 类似)

1首先可以由 \(x\) 在笛卡尔树的左儿子继承，即 \(i<x\)

2\(i=x\) 那么 \(f[l][x]=f[l][x-1]+h[x]\)

3\(i>x\) 那么 \(f[l][i]=min(f[l][x-1]+h[x]\times (i-x+1),f[x+1][i]+h[x]\times (x-l+1))\)

前两个好办，对于3，可以证明，这个 \(min\) 的函数是存在一个分界点，使得在左边一个更优，而右边另一个最优，并且具有可二分性

证明

只需要证明下面的东西即可

\(f[l][x-1]+(i-x+1)\times h[x]-(f[x+1][i]+(x-l+1)\times h[x])(\ge)\le\\
f[l][x-1]+((i+1)-x+1)\times h[x]-(f[x+1][i+1]+(x-l+1)\times h[x])\)

因为

\(f[x+1][i+1]-f[x+1][i]\le h[x]\)

那么可以证明上式取 \(\le\)

那么直接二分端点就好了

现在只需要支持区间加，区间对等差数列取 \(min\)，单点询问这些操作

可以离线线段树统计答案或者主席树在线回答

# include "meetings.h"
# include <bits/stdc++.h>
using namespace std;
typedef long long ll;

const int maxn(7.5e5 + 5);

struct Segment {
	ll covk[maxn << 2], covb[maxn << 2], add[maxn << 2], vl[maxn << 2], vr[maxn << 2];
	int cov[maxn << 2];

	inline void Cover(int x, int l, int r, ll k, ll b) {
		cov[x] = 1, add[x] = 0, vl[x] = k * l + b, vr[x] = k * r + b, covk[x] = k, covb[x] = b;
	}

	inline void Addv(int x, ll v) {
		add[x] += v, vl[x] += v, vr[x] += v;
	}

	inline void Pushdown(int x, int l, int mid, int r) {
		if (cov[x]) {
			Cover(x << 1, l, mid, covk[x], covb[x]);
			Cover(x << 1 | 1, mid + 1, r, covk[x], covb[x]);
			cov[x] = 0;
		}
		if (add[x]) Addv(x << 1, add[x]), Addv(x << 1 | 1, add[x]), add[x] = 0;
	}

	ll Query(int x, int l, int r, int p) {
		if (l == r) return vl[x];
		int mid = (l + r) >> 1;
		Pushdown(x, l, mid, r);
		ll ret = (p <= mid ? Query(x << 1, l, mid, p) : Query(x << 1 | 1, mid + 1, r, p));
		vl[x] = vl[x << 1], vr[x] = vr[x << 1 | 1];
		return ret;
	}

	void Chkmin(int x, int l, int r, int ql, int qr, ll k, ll b, ll v) {
		if (ql <= l && qr >= r) {
			if (k * r + b <= vr[x] + v && k * l + b <= vl[x] + v) {
			    Cover(x, l, r, k, b);
				return;
			}
			if (k * l + b >= vl[x] + v && k * r + b >= vr[x] + v) {
				Addv(x, v);
				return;
			}
		}
		int mid = (l + r) >> 1;
		Pushdown(x, l, mid, r);
		if (ql <= mid) Chkmin(x << 1, l, mid, ql, qr, k, b, v);
		if (qr > mid) Chkmin(x << 1 | 1, mid + 1, r, ql, qr, k, b, v);
		vl[x] = vl[x << 1], vr[x] = vr[x << 1 | 1];
	}
} fix_l, fix_r;

int q, n, h[maxn], st[20][maxn], lg[maxn], qryl[maxn], qryr[maxn];
vector <ll> ans;
vector <int> qry[maxn];

inline int Max(int x, int y) {
	return h[x] > h[y] ? x : y;
}

inline int RMQ(int l, int r) {
	int len = lg[r - l + 1];
	return Max(st[len][l], st[len][r - (1 << len) + 1]);
}

void Solve(int l, int r) {
	if (l > r) return;
	int mid = RMQ(l, r), i, len = qry[mid].size(), id;
	ll fl = 0, fr = 0;
	Solve(l, mid - 1), Solve(mid + 1, r);
	for (i = 0; i < len; ++i) {
		id = qry[mid][i];
		ans[id] = (ll)h[mid] * (qryr[id] - qryl[id] + 1);
		if (qryl[id] < mid) ans[id] = min(ans[id], (ll)(qryr[id] - mid + 1) * h[mid] + fix_r.Query(1, 1, n, qryl[id]));
		if (qryr[id] > mid) ans[id] = min(ans[id], (ll)(mid - qryl[id] + 1) * h[mid] + fix_l.Query(1, 1, n, qryr[id]));
	}
	if (l < mid) fl = fix_r.Query(1, 1, n, l);
	if (r > mid) fr = fix_l.Query(1, 1, n, r);
	fix_r.Chkmin(1, 1, n, l, mid, -h[mid], fr + (ll)h[mid] * (mid + 1), (ll)h[mid] * (r - mid + 1));
	fix_l.Chkmin(1, 1, n, mid, r, h[mid], fl - (ll)h[mid] * (mid - 1), (ll)h[mid] * (mid - l + 1));
}

vector <ll> minimum_costs(vector <int> _h, vector <int> _l, vector <int> _r) {
	int i, j, mid;
	q = _l.size(), n = _h.size();
	for (i = 0; i < n; ++i) st[0][i + 1] = i + 1, h[i + 1] = _h[i];
	for (i = 2; i <= n; ++i) lg[i] = lg[i >> 1] + 1;
	for (j = 1; j <= lg[n]; ++j)
		for (i = 1; i + (1 << j) - 1 <= n; ++i)
			st[j][i] = Max(st[j - 1][i], st[j - 1][i + (1 << (j - 1))]);
	ans.resize(q);
	for (i = 0; i < q; ++i) {
		mid = RMQ(_l[i] + 1, _r[i] + 1);
		qry[mid].push_back(i);
		qryl[i] = _l[i] + 1, qryr[i] = _r[i] + 1;
	}
	Solve(1, n);
	return ans;
}