题解：[ZJOI2007] 仓库建设

题目传送门

朴素 DP

记 \(\textit{dp}_i\) 表示第 \(i\) 个工厂必须建仓库，第 \(1\sim i\) 个工厂的总花费。

则有：

\[\begin{aligned} \textit{dp}_i&=c_i+\min_{j=0}^{i-1}\left(\textit{dp}_j+\sum_{k=j+1}^i(x_i-x_k)p_k\right)\\ &=c_i+\min_{j=0}^{i-1}\left(\textit{dp}_j+x_i\sum_{k=j+1}^ip_k-\sum_{k=j+1}^{i-1}x_kp_k\right) \end{aligned} \]

记：

\[\begin{aligned} \textit{prep}_i&=\sum_{j=1}^ip_j\\ \textit{prexp}_i&=\sum_{j=1}^ix_jp_j \end{aligned} \]

则有：

\[\begin{aligned} \textit{dp}_i&=c_i+\min_{j=0}^{i-1}\left(\textit{dp}_j+\sum_{k=j+1}^i(x_i-x_k)p_k\right)\\ &=c_i+\min_{j=0}^{i-1}\left(\textit{dp}_j+x_i\left(\textit{prep}_i-\textit{prep}_j\right)-\left(\textit{prexp}_i-\textit{prexp}_j\right)\right)\\ &=c_i+x_i\textit{prep}_i-\textit{prexp}_i+\min_{j=0}^{i-1}\left(\textit{dp}_j-x_i\textit{prep}_j+\textit{prexp}_j\right) \end{aligned} \]

时间复杂度：\(\mathcal O\left(n^2\right)\)。

斜率优化 DP

发现递推式中关于 \(j\) 的部分 \(\textit{dp}_j-x_i\textit{prep}_j+\textit{prexp}_j\) 仅有一项与 \(i\) 相关，考虑斜率优化 DP。

设决策点 \(j_1<j_2\)，钦定 \(j_1\) 优于 \(j_2\)，有：

\[\begin{aligned} \textit{dp}_{j_1}-x_i\textit{prep}_{j_1}+\textit{prexp}_{j_1}&\leq \textit{dp}_{j_2}-x_i\textit{prep}_{j_2}+\textit{prexp}_{j_2}\\ x_i(\textit{prep}_{j_2}-\textit{prep}_{j_1})&\leq\left(\textit{dp}_{j_2}+\textit{prexp}_{j_2}\right)-\left(\textit{dp}_{j_1}+\textit{prexp}_{j_1}\right)\\ x_i&\leq\dfrac{\left(\textit{dp}_{j_2}+\textit{prexp}_{j_2}\right)-\left(\textit{dp}_{j_1}+\textit{prexp}_{j_1}\right)}{\textit{prep}_{j_2}-\textit{prep}_{j_1}}\\ x_i&\leq\dfrac{\left(\textit{dp}_{j_1}+\textit{prexp}_{j_1}\right)-\left(\textit{dp}_{j_2}+\textit{prexp}_{j_2}\right)}{\textit{prep}_{j_1}-\textit{prep}_{j_2}}\\ \end{aligned} \]

因此可以有决策点 \(A_j\left(\textit{prep}_j,\textit{dp}_j+\textit{prexp}_j\right)\)。单调队列维护下凸壳即可。

时间复杂度 \(\mathcal O(n)\)。

AC 代码

注意 double 精度问题。

//#include<bits/stdc++.h>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<iomanip>
#include<cstdio>
#include<string>
#include<vector>
#include<cmath>
#include<ctime>
#include<deque>
#include<queue>
#include<stack>
#include<list>
using namespace std;
typedef long long ll;
constexpr const int N=1000000;
constexpr const ll inf=0x3f3f3f3f3f3f3f3f;
constexpr const double eps=1e-8;
int n,x[N+1],p[N+1],c[N+1];
ll prep[N+1],prexp[N+1],dp[N+1];
struct slopeDP{
	struct node{
		ll x,y;
		int id;
	}q[N+1];
	int front,rear;
	slopeDP(){
		front=1;rear=0;
	}
	double slope(node i,node j){
		return 1.0*(i.y-j.y)/(i.x-j.x);
	}
	bool check(node a,node b,node c,node d){
		return (a.y-b.y)*(c.x-d.x) < (c.y-d.y)*(a.x-b.x);
	}
	void push(int id){
		node x={prep[id],dp[id]+prexp[id],id};
		//slope(q[rear],x)<slope(q[rear-1],q[rear])
		while(front+1<=rear && check(q[rear],x,q[rear-1],q[rear])){
			rear--;
		}
		q[++rear]=x;
	}
	int query(int x){
		while(front+1<=rear && slope(q[front],q[front+1])<x){
			front++;
		}
		return q[front].id;
	}
}t;
int main(){
	/*freopen("test.in","r",stdin);
	freopen("test.out","w",stdout);*/
	
	ios::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);
	
	cin>>n;
	for(int i=1;i<=n;i++){
		cin>>x[i]>>p[i]>>c[i];
		prep[i]=prep[i-1]+p[i];
		prexp[i]=prexp[i-1]+1ll*x[i]*p[i];
	}
	t.push(0);
	for(int i=1;i<=n;i++){
		int j=t.query(x[i]);
		dp[i]=c[i]+1ll*x[i]*prep[i]-prexp[i]+dp[j]-x[i]*prep[j]+prexp[j];
		if(!p[i]){
			dp[i]=min(dp[i],dp[i-1]);
		}
		t.push(i);
	}
	cout<<dp[n]<<'\n';
	
	cout.flush();
	
	/*fclose(stdin);
	fclose(stdout);*/
	return 0;
}
/*
3
0 5 10
5 3 100
9 6 10

32
*/