感覺和鋸木廠那個題很類似的。

其實這個題還那個題唯一的區別就是\(dp\)轉移式子中的\(f\)變成了\(g\)

qwq不想多說了

直接看我的前一篇題解吧qwq

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<queue>
#include<map>
#include<set>
#define mk make_pair
#define ll long long
#define int long long
using namespace std;
inline int read()
{
  int x=0,f=1;char ch=getchar();
  while (!isdigit(ch)) {if (ch=='-') f=-1;ch=getchar();}
  while (isdigit(ch)) {x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
  return x*f;
}
const int maxn = 1e6+1e2;
int n,m;
int w[maxn],d[maxn];
int f[maxn];
int dis[maxn],val[maxn],sum[maxn];
int sw[maxn];
struct Point{
    int x,y,num;
};
Point q[maxn];
int chacheng(Point x,Point y)
{
    return x.x*y.y-x.y*y.x;
}
bool count(Point i,Point j,Point k)
{
    Point x,y;
    x.x=k.x-i.x;
    x.y=k.y-i.y;
    y.x=k.x-j.x;
    y.y=k.y-j.y;
    if (chacheng(x,y)<=0) return true;
    return false;

int head=1,tail=0;
void push(Point x)
{
    while (tail>=head+1 && count(q[tail-1],q[tail],x)) tail--;
    q[++tail]=x;
}
void pop(int lim)
{
    while(tail>=head+1 && q[head+1].y-q[head].y < lim * (q[head+1].x-q[head].x)) head++;
}
int g[maxn];
int ymh[maxn];
signed main()
{
  n=read();
  for (int i=1;i<=n;i++) d[i]=read(),w[i]=read(),ymh[i]=read();
  for (int i=1;i<=n;i++) dis[i]=d[n]-d[i];
  for (int i=1;i<=n;i++) sw[i]=sw[i-1]+w[i];
  for (int i=1;i<=n;i++) val[i]=w[i]*dis[i];
  for (int i=1;i<=n;i++) sum[i]=sum[i-1]+val[i];
  g[0]=sum[n];
  push((Point){0,sum[n],0});
  for (int i=1;i<=n;i++)
  {
     pop((-1)*dis[i]);
     int now = q[head].num;
     g[i]=min(g[now]-(sw[i]-sw[now])*dis[i]+ymh[i],g[i-1]);
     push((Point){sw[i],g[i],i});
  } 
  cout<<g[n]+ymh[n];
  return 0;
}