呃，不难，什么方程都在注释里面。

<span style="font-family:KaiTi_GB2312;font-size:18px;">#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define N 1001000
#define inf 0x3f3f3f3f
/*
	f[i]=f[j]+sum[i]-sum[j]-p[j]*(x[i]-x[j])+c[i];

	f[j]-sum[j]+p[j]*x[j]=x[i]*p[j]+f[i]-sum[i]-c[i]
	y=f[j]-sum[j]+p[j]*x[j]
	x=p[j]
	k=x[i]
	b=f[i]-sum[i]-c[i]
*/
using namespace std;
int n,x[N],c[N];
long long p[N],sum[N],f[N];
struct Point
{
	long long x,y;
	int id;
	Point(long long _x,long long _y,int _id):x(_x),y(_y),id(_id){}
	Point(){}
}now,q[N];
int l,r;
inline long long xmul(Point i,Point j,Point k){return (i.y-j.y)*(j.x-k.x)-(j.y-k.y)*(i.x-j.x);}
int main()
{
//	freopen("test.in","r",stdin);
	scanf("%d",&n);
	memset(f,0x3f,sizeof(f));
	f[0]=0;
	//初始点经过计算为(0,0)
	for(int i=1;i<=n;i++)
	{
		scanf("%d%d%d",&x[i],&p[i],&c[i]);
		p[i]+=p[i-1];
		sum[i]=sum[i-1]+p[i-1]*(x[i]-x[i-1]);

		int K=x[i];
		while(l<r&&q[l].y-q[l+1].y>=(q[l].x-q[l+1].x)*K)l++; //斜率小于K就跳
		int u=q[l].id;
		f[i]=f[u]+sum[i]-sum[u]-p[u]*(x[i]-x[u])+c[i];
		now=Point(p[i],f[i]-sum[i]+p[i]*x[i],i);

		while(l<r&&xmul(now,q[r],q[r-1])<=0)r--; //斜(i,j)<=斜(j,k)就弹
		q[++r]=now;
	}
	cout<<f[n]<<endl;
	return 0;
}
</span>

【1096】【ZJOI2007】仓库建设 斜率优化DP