凸包模板

1.构造一个完整凸包

2.普通的构造一次凸包并求答案

2.允许动态加点的凸包求答案（cf932f）

//#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")

//#include <immintrin.h>
//#include <emmintrin.h>
#include <bits/stdc++.h>
using namespace std;
#define rep(i,h,t) for (int i=h;i<=t;i++)
#define dep(i,t,h) for (int i=t;i>=h;i--)
#define ll long long
#define me(x) memset(x,0,sizeof(x))
#define IL inline
#define rint register int
inline ll rd(){
    ll x=0;char c=getchar();bool f=0;
    while(!isdigit(c)){if(c=='-')f=1;c=getchar();}
    while(isdigit(c)){x=(x<<1)+(x<<3)+(c^48);c=getchar();}
    return f?-x:x;
}
char ss[1<<24],*A=ss,*B=ss;
IL char gc()
{
    return A==B&&(B=(A=ss)+fread(ss,1,1<<24,stdin),A==B)?EOF:*A++;
}
template<class T>void maxa(T &x,T y)
{
    if (y>x) x=y;
}
template<class T>void mina(T &x,T y)
{
    if (y<x) x=y;
}
template<class T>void read(T &x)
{
    int f=1,c; while (c=gc(),c<48||c>57) if (c=='-') f=-1; x=(c^48);
    while(c=gc(),c>47&&c<58) x=x*10+(c^48); x*=f;
}
const int mo=1e9+7;
ll fsp(int x,int y)
{
    if (y==1) return x;
    ll ans=fsp(x,y/2);
    ans=ans*ans%mo;
    if (y%2==1) ans=ans*x%mo;
    return ans;
}
struct cp {
    ll x,y;
    cp operator +(cp B)
    {
        return (cp){x+B.x,y+B.y};
    }
    cp operator -(cp B)
    {
        return (cp){x-B.x,y-B.y};
    }
    ll operator *(cp B)
    {
        return x*B.y-y*B.x;
    }
    int half() { return y < 0 || (y == 0 && x < 0); }
};
int QF;
struct line{
    mutable ll x,y,k;
    bool operator <(const line o) const{
        return !QF?x<o.x:k<o.k;
    }
};
const int N=2e5;
struct linec:multiset<line>{
  ll div(ll a,ll b)
  {
      return a/b-((a^b)<0&&a%b);
  }
  bool xj(auto x,auto y)
  {
      if (y==end()) { x->k=1e18; return 0;}
      if (x->x==y->x) {x->k=y->y>x->y?1e18:-1e18;}
      else x->k=div(y->y-x->y,y->x-x->x);
      return x->k>=y->k;
  }
  void add(ll xx,ll yy)
  {
      auto z=insert({xx,yy,0}),y=z++,x=y;
      while (xj(y,z)) z=erase(z);
      if (x!=begin()&&xj(--x,y)) xj(x,y=erase(y));
      while ((y=x)!=begin()&&(--x)->k>=y->k) xj(x,erase(y));
  }
  ll query(ll x)
  {
       QF=1;
       auto l=*lower_bound({0,0,x});
       QF=0;
       return -l.x*x+l.y;
  }
}S[N];
struct re{
    int a,b,c;
};
int a[N],b[N];
vector<int> ve[N];
int pos[N],size[N],cnt;
ll dp[N];
void dfs(int x,int y)
{
    size[x]=1;
    int ans=0;
    for(auto v:ve[x])
    if (v!=y) 
    {
        dfs(v,x);
        size[x]+=size[v];
        if (size[v]>size[ans]) ans=v;
    }
    if (ans)
    {
        pos[x]=pos[ans];
        for(auto v:ve[x])
        if (v!=y&&v!=ans)
        {
            for (auto u:S[pos[v]])
              S[pos[x]].add(u.x,u.y); 
        }
        dp[x]=S[pos[x]].query(a[x]);
    } else
    {
        pos[x]=++cnt;
    }
    S[pos[x]].add(-b[x],dp[x]); 
}
int main()
{
   freopen("1.in","r",stdin);
   freopen("1.out","w",stdout);
   ios::sync_with_stdio(false);
   int n;
   cin>>n;
   rep(i,1,n) cin>>a[i];
   rep(i,1,n) cin>>b[i];
   rep(i,1,n-1)
   {
        int x,y;
        cin>>x>>y;
        ve[x].push_back(y);
        ve[y].push_back(x);
   }
   dfs(1,0);
   rep(i,1,n-1) cout<<dp[i]<<" ";
   cout<<dp[n]<<endl;
   return 0;
}

这个板子非常短而且挺好用的

mutable能使const也能被修改