斜率优化

upd: 2021/07/20

学到一个简单点的方法。
还是以Print Article 为例。
给出 \(n\)， \(m\) 和数列，将 \(n\) 个数分成任意段，每一段贡献是 \(sum^2+m\)，求最小总贡献
\(f_i = min(f_j + m + {(s_i-s_j)}^2)\)

\[f_i = min(f_j + m + {(s_i-s_j)}^2)\\ f_i = f_j + m + s^2_i + s^2_j -2s_is_j\\ f_i - m - s^2_i = -2s_i s_j + f_j + s^2_j\\ 令 x = s_j, k = 2 s_i, y = f_j + s^2_j, b = f_i - m -s^2_i\\ b = -kx + y\\ y = kx + b\\ k 从 1\sim n一直递增\\ 想要使得 b 尽量小，如图，维护凸包，寻找切线\\ \]

---

旧：devinwang的斜率优化入门题单简要题解
大张旗鼓开了个博客结果博客主要内容是“常规操作，过”TAT
板子

int h = 0, t = -1; q[++t] = 0;
for(int i = 1; i <= n; i++){
	while(h < t && slope(q[h + 1], q[h]) <= i) h++;
	f[i] = f[q[h]] + a[i] + i * (s[i] - s[q[h]]);
	while(h < t && slope(i, q[t]) <= slope(q[t], q[t - 1])) t--;
	q[++t] = i; 
}

Print Article

题意

给出 \(n\)， \(m\) 和数列

将 \(n\) 个数分成任意段，每一段贡献是 \(sum^2+m\)，求最小总贡献

题解

当 \(k < j < i\) 时，如果 \(j\) 优于 \(k\)，那么

\[f_j + m + {(s_i-s_j)}^2 \le f_k + m + {(s_i - s_k)}^2\\ (f_j+s^2_j) - (f_k + s^2_k) \le 2(s_j - s_k) *s_i\\ \frac {(f_j+s^2_j) - (f_k + s^2_k)} {2(s_j - s_k)} \le s_i \]

不妨 \(y(i) = f_i + s^2_i, x(i) = 2 s_i\)

则 \(\frac {y(j) - y(k)} {x(j) - x(k)} \le s_i\)

即 斜率 \(k \le s_i\)

令 \(g(j, k) = \frac {(f_j+s^2_j) - (f_k + s^2_k)} {2(s_j - s_k)}\)，

当 \(g(i, j) \le g(j, k)\) 时，

若 \(g(i, j) \le s_i\) 则 \(i\) 优于 \(j\) ，\(j\) 没有存在必要

若 \(g(i, j) > s_i\) 则 \(j\) 优于 \(i\) ，同样，\(g(j,k) > s_i\) ， \(k\) 优于 \(j\) ， \(j\) 没有存在必要

因此，剔除所有 \(g(i, j) \le g(j, k)\)，维护一个类似凸包的东西。

代码

const int N = 500010;
int n, m, f[N], s[N], q[N];
int gety(int j, int k){
	return f[j] + s[j] * s[j] - f[k] - s[k] * s[k];
}
int getx(int j, int k){
	return 2 * (s[j] - s[k]);
}
int main(){
	while(scanf("%d%d", &n, &m) == 2){
		for(int i = 1; i <= n; i++)
			scanf("%d", &s[i]), s[i] += s[i - 1];
		int h = 0, t = -1; q[++t] = 0;
		for(int i = 1; i <= n; i++){
			while(h < t && gety(q[h + 1], q[h]) <= s[i] * getx(q[h + 1], q[h])) h++;
			f[i] = f[q[h]] + (s[i] - s[q[h]]) * (s[i] - s[q[h]]) + m;
			while(h < t && gety(i, q[t]) * getx(q[t], q[t - 1]) <= getx(i, q[t]) * gety(q[t], q[t - 1])) t--;
			q[++t] = i; 
		}
		printf("%d\n", f[n]);
	}
	return 0;
}

用slope的写法：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define mkp make_pair
#define pb push_back
#define PII pair<int, int>
#define PLL pair<ll, ll>
#define ls(x) ((x) << 1)
#define rs(x) ((x) << 1 | 1)
#define fi first
#define se second
const int N = 500010;
int n, m, q[N];
ll f[N], s[N];
ll getx(int x, int y) {
	return s[y] - s[x];
}
ll gety(int x, int y) {
	return f[y] + s[y] * s[y] - f[x] - s[x] * s[x];
}
double slope(int x, int y) {
	if(getx(x, y) == 0) {
		return (gety(x, y) >= 0) ? 1 : -1;
	}
	return 1.0 * gety(x, y) / getx(x, y);
}
int main(){
	while(scanf("%d%d", &n, &m) == 2) {
		for(int i = 1; i <= n; i++)
			scanf("%lld", &s[i]), s[i] += s[i - 1];
		int h = 1, t = 0; q[++t] = 0;
		for(int i = 1; i <= n; i++) {
			while(h < t && slope(q[h], q[h + 1]) <= 2 * s[i]) h++;
			f[i] = f[q[h]] + (s[i] - s[q[h]]) * (s[i] - s[q[h]]) + m;
			while(h < t && slope(q[t - 1], q[t]) >= slope(q[t], i)) t--;
			q[++t] = i;
		} 
		printf("%lld\n", f[n]);
	}
	
	return 0;
}
/*
f[i]-s[i]^2-m=f[j]+s[j]^2-2*s[i]*s[j]
k=2*s[i],x=s[j]
y=f[j]+s[j]^2
b=f[i]-s[i]^2-m
b=y-kx
y=kx+b
k从1~n单调不降
 
*/

玩具装箱

同Print Article，

\[f_j+ {(j + s_j)} ^ 2 - f_k - {(k+ s_k)}^2 < 2(i + s_i - L - 1) * (j + s_k - k - s_k) \]

const int N = 500010;
int n, q[N];
ll f[N], s[N], L;
ll gety(int j, int k){
	return f[j] + (j + s[j]) * (j + s[j]) - f[k] - (k + s[k]) * (k + s[k]);
}
ll getx(int j, int k){
	return 2ll * (s[j] + j - s[k] - k);
}
int main(){
	scanf("%d%lld", &n, &L);
	for(int i = 1; i <= n; i++)
		scanf("%lld", &s[i]), s[i] += s[i - 1];
	int h = 0, t = -1; q[++t] = 0;
	for(int i = 1; i <= n; i++){
		while(h < t && gety(q[h + 1], q[h]) <= (i + s[i] - L - 1) * getx(q[h + 1], q[h])) h++;
		f[i] = f[q[h]] + (i - q[h] + s[i] - s[q[h]] - L - 1) * (i - q[h] + s[i] - s[q[h]] - L - 1);
		while(h < t && gety(i, q[t]) * getx(q[t], q[t - 1]) <= getx(i, q[t]) * gety(q[t], q[t - 1])) t--;
		q[++t] = i; 
	}
	printf("%lld\n", f[n]);
	return 0;
}

用slope的写法

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define mkp make_pair
#define pb push_back
#define PII pair<int, int>
#define PLL pair<ll, ll>
#define ls(x) ((x) << 1)
#define rs(x) ((x) << 1 | 1)
#define fi first
#define se second
const int N = 500010;
int n, m, q[N];
ll f[N], s[N];
ll getx(int x, int y) {
	return y + s[y] - (x + s[x]);
}
ll gety(int x, int y) {
	return f[y] + (y + s[y] + m) * (y + s[y] + m) - f[x] - (x + s[x] + m) * (x + s[x] + m);
}
double slope(int x, int y) {
//	if(getx(x, y) == 0) {
//		return (gety(x, y) >= 0) ? 1 : -1;
//	}
	return 1.0 * gety(x, y) / getx(x, y);
}
int main(){
	scanf("%d%d", &n, &m);
	for(int i = 1; i <= n; i++)
		scanf("%lld", &s[i]), s[i] += s[i - 1];
	int h = 1, t = 0; q[++t] = 0;
	for(int i = 1; i <= n; i++) {
		while(h < t && slope(q[h], q[h + 1]) <= 2 * (i + s[i])) h++;
		f[i] = f[q[h]] + (i - q[h] - 1 + s[i] - s[q[h]] - m) * (i - q[h] - 1 + s[i] - s[q[h]] - m);
		while(h < t && slope(q[t - 1], q[t]) >= slope(q[t], i)) t--;
		q[++t] = i;
	} 
	printf("%lld\n", f[n]);
	return 0;
}=

锯木厂选址

题意

从山顶上到山底下沿着一条直线种植了 \(n\) 棵老树。当地的政府决定把他们砍下来。为了不浪费任何一棵木材，树被砍倒后要运送到锯木厂。
木材只能朝山下运。山脚下有一个锯木厂。另外两个锯木厂将新修建在山路上。你必须决定在哪里修建这两个锯木厂，使得运输的费用总和最小。假定运输每公斤木材每米需要一分钱。
你的任务是编写一个程序，读入树的个数和他们的重量与位置，计算最小运输费用。

懒得翻译（（（

题解

考虑总的贡献，山顶到山脚从小到大读入，假设 \(j < i\)，

令 \(d_j\) 为到山脚的距离。

选 \(i\) 的时候，最小总贡献

\[f_i = min(\sum_{x = 1} ^ j (d_x-d_j)*w_x + \sum_{x = j + 1} ^ {i} (d_x - d_i) * w_x + \sum_{x = i + 1}^n d_x*w_x) \]

令 \(s_i\) 为 \(w_i\) 的前缀和

\[f_i = \sum_{x = 1}^n d_x*w_x-\sum_{x = 1} ^ j d_j*w_x -\sum_{x = j + 1} ^ {i} d_i * w_x\\ f_i = SUM-d_j*s_j -d_i (s_i - s_j)\\ f_i = SUM-max(d_j*s_j +d_i (s_i - s_j))\\ \]

当 \(k < j\) 且 \(j\) 优于 \(k\)

\[d_j s_j + d_i(s_i - s_j) > d_k * s_k + d_i (s_i - s_k)\\ \frac {d_js_j-d_ks_k} {s_j - s_k} > d_i \]

当 \(slope(i, j) > slope(j, k)\) 时 \(j\) 无用。

维护一个下凸包。

代码

const int N = 2e5 + 10;
int n, q[N];
ll s[N], w[N], d[N], sum;
ll gety(int j, int k){
	return d[j] * s[j] - d[k] * s[k];
}
ll getx(int j, int k){
	return s[j] - s[k];
}
double slope(int j, int k){
	return 1.0 * gety(j, k) / getx(j, k);
}
ll calc(int j, int i){
	return sum - d[j] * s[j] - d[i] * (s[i] - s[j]);
}
int main(){
	scanf("%d", &n);
	for(int i = 1; i <= n; i++)
		scanf("%lld%lld", &w[i], &d[i]), s[i] = s[i - 1] + w[i];
	for(int i = n; i >= 1; i--)
		d[i] += d[i + 1], sum += d[i] * w[i];
		
	int h = 0, t = -1; q[++t] = 0;
	ll ans = 1e18;
	for(int i = 1; i <= n; i++){
		while(h < t && slope(q[h + 1], q[h]) > d[i]) h++;
		ans = min(ans, calc(q[h], i));
		while(h < t && slope(i, q[t]) > slope(q[t], q[t - 1])) t--;
		q[++t] = i; 
	}
	printf("%lld\n", ans);
	return 0;
}

仓库建设

和前一题差不多吧（（（

const int N = 1e6 + 10;
int n, q[N];
ll s[N], w[N], d[N], g[N], f[N], c[N];
ll gety(int j, int k){
	return f[j] + g[j] - f[k] - g[k];
}
ll getx(int j, int k){
	return s[j] - s[k];
}
double slope(int j, int k){
	return 1.0 * gety(j, k) / getx(j, k);
}
int main(){
	scanf("%d", &n);
	for(int i = 1; i <= n; i++)
		scanf("%lld%lld%lld", &d[i], &w[i], &c[i]), 
		s[i] = s[i - 1] + w[i], g[i] = g[i - 1] + d[i] * w[i]; 
		
	int h = 0, t = -1; q[++t] = 0;
	ll ans = 1e18;
	for(int i = 1; i <= n; i++){
		while(h < t && slope(q[h + 1], q[h]) <= d[i]) h++;
		f[i] = f[q[h]] + c[i] + (s[i] - s[q[h]]) * d[i] - g[i] + g[q[h]]; 
		while(h < t && slope(i, q[t]) <= slope(q[t], q[t - 1])) t--;
		q[++t] = i; 
	}
	printf("%lld\n", f[n]);
	return 0;
}

土地购买

题解

按照从小到大排序，\(a\) 第一关键字，\(b\)第二关键字，去除包含。

现在 \(a\) 始终递增，\(b\) 始终递减。

显然取连续的一段才是最优的。

然后常规操作

代码

const int N = 1e6 + 10;
int n, q[N];
ll f[N];
struct node{
	ll a, b;
	bool operator < (const node x) const {
		return (a == x.a) ? b < x.b : a < x.a;
	}
}po[N];
ll gety(int j, int k){
	return f[j] - f[k];
}
ll getx(int j, int k){
	return po[k + 1].b - po[j + 1].b;
}
double slope(int j, int k){
	return 1.0 * gety(j, k) / getx(j, k);
}
int main(){
	scanf("%d", &n);
	for(int i = 1; i <= n; i++)
		scanf("%lld%lld", &po[i].a, &po[i].b);
	sort(po + 1, po + n + 1);
	int cnt = 0;
	for(int i = 1; i <= n; i++){
		while(cnt && po[cnt].b <= po[i].b) cnt--;
		po[++cnt] = po[i];
	}
	
	int h = 0, t = -1; q[++t] = 0;
	ll ans = 1e18;
	for(int i = 1; i <= cnt; i++){
		while(h < t && slope(q[h + 1], q[h]) <= po[i].a) h++;
		f[i] = f[q[h]] + po[i].a * po[q[h] + 1].b; 
		while(h < t && slope(i, q[t]) <= slope(q[t], q[t - 1])) t--;
		q[++t] = i; 
	}
	printf("%lld\n", f[cnt]);
	return 0;
}

特别行动队

啊这就常规操作

然后发现常数项最好拖到主函数那里面，否则会炸精度（（（

const int N = 1e6 + 10;
int n, q[N];
ll f[N], s[N], a, b, c;
ll gety(int j, int k){
	return f[j] + a * s[j] * s[j] - b * s[j] - f[k] - a * s[k] * s[k] + b * s[k];
}
ll getx(int j, int k){
	return s[j] - s[k];
}
double slope(int j, int k){
	return 1.0 * gety(j, k) / getx(j, k);
}
int main(){
	scanf("%d", &n);
	scanf("%lld%lld%lld", &a, &b, &c);
	for(int i = 1; i <= n; i++)
		scanf("%lld", &s[i]), s[i] += s[i - 1];
	int h = 0, t = -1; q[++t] = 0;
	for(int i = 1; i <= n; i++){
		while(h < t && slope(q[h + 1], q[h]) > 2 * a * s[i]) h++;
		ll x = s[i] - s[q[h]]; f[i] = f[q[h]] + a * x * x + b * x + c; 
		while(h < t && slope(i, q[t]) > slope(q[t], q[t - 1])) t--;
		q[++t] = i; 
	}
	printf("%lld\n", f[n]);
	return 0;
}

序列分割

题解

发现切的顺序对答案无影响

然后常规操作

注意到 \(s_j = s_k\) 时，要特判，否则会除以0 RE

代码

const int N = 1e5 + 10;
int n, k, q[N];
ll f[N], s[N], g[N], pre[210][N];
ll gety(int j, int k){
	return g[j] - s[j] * s[j] - (g[k] - s[k] * s[k]);
}
ll getx(int j, int k){
	return s[k] - s[j];
}
double slope(int j, int k){
	if(s[j] == s[k]) return -1e18;
	return 1.0 * gety(j, k) / getx(j, k);
}
int main(){
	scanf("%d%d", &n, &k);
	for(int i = 1; i <= n; i++)
		scanf("%lld", &s[i]), s[i] += s[i - 1];
	for(int j = 1; j <= k; j++){
		int h = 0, t = -1; q[++t] = 0;
		for(int i = 1; i <= n; i++) g[i] = f[i], f[i] = 0;
		for(int i = 1; i <= n; i++){
			while(h < t && slope(q[h + 1], q[h]) <= s[i]) h++;
			f[i] = g[q[h]] + s[q[h]] * (s[i] - s[q[h]]); pre[j][i] = q[h];
		//	cout<<i<<"*"<<q[h]<<endl; 
			while(h < t && slope(i, q[t]) <= slope(q[t], q[t - 1])) t--;
			q[++t] = i; 
		}
	}
	printf("%lld\n", f[n]);
	int j = k, i = n;
	while(j) {
		printf("%d ", pre[j][i]);
		i = pre[j][i], j--;
	}
	return 0;
}

「SDOI2016」征途

题解

一通推柿子答案为

\[min(m\sum x^2 - {(\sum x)}^2) \]

一通常规操作

注意代码中注释的那行，0是显然错误的

代码

const int N = 3010 + 10;
int n, q[N];
ll f[N], g[N], s[N] ,m;
ll gety(int j, int k){
	return g[j] + s[j] * s[j] - (g[k] + s[k] * s[k]);
}
ll getx(int j, int k){
	return s[j] - s[k];
}
double slope(int j, int k){
	return 1.0 * gety(j, k) / getx(j, k);
}
int main(){
	scanf("%d%lld", &n, &m);
	for(int i = 1; i <= n; i++)
		scanf("%lld", &s[i]), s[i] += s[i - 1], f[i] = s[i] * s[i];
	for(int j = 1; j < m; j++){
		int h = 0, t = -1; q[++t] = j;//q[++t] = 0;
		for(int i = 1; i <= n; i++) g[i] = f[i], f[i] = 0;
		for(int i = j + 1; i <= n; i++){
			while(h < t && slope(q[h + 1], q[h]) <= 2 * s[i]) h++;
			ll x = s[i] - s[q[h]]; f[i] = g[q[h]] + x * x;
			while(h < t && slope(i, q[t]) <= slope(q[t], q[t - 1])) t--;
			q[++t] = i; 
		}
	}
	printf("%lld\n", - s[n] * s[n] + m * f[n]);
	return 0;
}

小P的牧场

常规操作。。。

const int N = 1e6 + 10;
int n, q[N];
ll a[N], b[N], f[N], s[N], sum;
ll gety(int j, int k){
	return f[j] - f[k];
}
ll getx(int j, int k){
	return s[j] - s[k];
}
double slope(int j, int k){
	return 1.0 * gety(j, k) / getx(j, k);
}
int main(){
	scanf("%d%", &n);
	for(int i = 1; i <= n; i++)
		scanf("%lld", &a[i]);
	for(int i = 1; i <= n; i++)
		scanf("%lld", &b[i]),
		s[i] = s[i - 1] + b[i], sum += i * b[i];
	int h = 0, t = -1; q[++t] = 0;
	for(int i = 1; i <= n; i++){
		while(h < t && slope(q[h + 1], q[h]) <= i) h++;
		f[i] = f[q[h]] + a[i] + i * (s[i] - s[q[h]]);
		while(h < t && slope(i, q[t]) <= slope(q[t], q[t - 1])) t--;
		q[++t] = i; 
	}
	printf("%lld\n", f[n] - sum);
	return 0;
}
/*
4
2 4 2 4
3 1 4 2
*/