[ZJOI2007]仓库建设

这题斜率优化DP即可。
一如既往地推公式：
设s1[i]=p[1~i]的和，s2[i]=sigma(p[k]*x[k])(1<=k<=i)

f[i]=min{f[j]+x[i]*(s1[i-1]-s1[j])-s2[i-1]+s2[j]}+c[i]

设k<j<i且j比k更优。

(f[j]+s2[j])-(f[k]+s2[k])/(s1[j]-s1[k])<x[i]

上标：

#include<cstdio>
#define ll long long
#define db double
#define N 1000010
using namespace std;
int n,g[N],len=0,l=0;
ll s1[N],s2[N],f[N],x[N],p[N],c[N];

inline int read()
{
	int x=0; char c=getchar();
	while (c<'0' || c>'9') c=getchar();
	while (c>='0' && c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
	return x;
}

db solve(int x,int y) {return (db)(f[x]+s2[x]-f[y]-s2[y])/(s1[x]-s1[y]);}

int main()
{
	freopen("P2120.in","r",stdin);
	freopen("P2120.out","w",stdout);
//	f[i]=min{f[j]+x[i]*(s1[i-1]-s1[j])-s2[i-1]+s2[j]}+c[i]
//	(f[j]+s2[j])-(f[k]+s2[k])/(s1[j]-s1[k])<x[i]
	n=read();
	for (int i=1;i<=n;i++)
	{
		x[i]=read(),p[i]=read(),c[i]=read();
		s1[i]=s1[i-1]+p[i],s2[i]=s2[i-1]+p[i]*x[i];
	}
	for (int i=1;i<=n;i++)
	{
		while (l<len && solve(g[l+1],g[l])<x[i]) l++;
		f[i]=f[g[l]]+x[i]*(s1[i-1]-s1[g[l]])-s2[i-1]+s2[g[l]]+c[i];
		while (solve(g[len],g[len-1])>=solve(i,g[len]) && len) len--;
		g[++len]=i;
	}
	printf("%lld
",f[n]);
	return 0;
}