Gym

题意：给点N棵树，前K棵是已经拥有的，现在可以再拥有一棵树，问形成的最大凸包面积。

思路：先求K棵树的凸包C，然后对于后面的N-K棵树，我们先判断是否在凸包内，如果不在，我们要求两个切线。 这里分类讨论，即可。

如果点在C的左边，那么两条切线分别一上一下； 如果在下边，两条切线一左一右。 然后去对应区间二分即可。 

（好像还有双指针的线性做法：求两个凸包，维护两条切线即可。

#include<bits/stdc++.h>
#define ll long long
#define rep(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
const int maxn=200010;
struct point{
    ll x,y;
    point(){}
    point(ll xx,ll yy):x(xx),y(yy){}
};
bool cmp(point w,point v){
    if(w.x!=v.x) return w.x<v.x;
    return w.y<v.y;
}
ll det(point a,point b){ return a.x*b.y-a.y*b.x;}
ll dot(point a,point b){ return a.x*b.x+a.y*b.y;}
point operator +(point a,point b){ return point(a.x+b.x,a.y+b.y);}
point operator -(point a,point b){ return point(a.x-b.x,a.y-b.y);}
point a[maxn],ch[maxn]; int top,ttop;
void convexhull(int N)
{
    for(int i=1;i<=N;i++){
        while(top>1&&det(ch[top]-ch[top-1],a[i]-ch[top-1])<=0) top--;
        ch[++top]=a[i];
    }
    ttop=top;
    for(int i=N-1;i>=1;i--){
        while(top>ttop&&det(ch[top]-ch[top-1],a[i]-ch[top-1])<=0) top--;
        ch[++top]=a[i];
    }
}
int get(int L,int R,int i,int w)
{
    while(L<R){
        int Mid=(L+R)>>1;
        if(det(ch[Mid]-a[i],ch[Mid+1]-a[i])*w>0) R=Mid;
        else L=Mid+1;
    }
    return L;
}
int bord(int L,int R,int i,int w)
{
    while(L<R){
        int Mid=(L+R)>>1;
        if((ch[Mid].x-a[i].x)*w<0) L=Mid+1;
        else R=Mid;
    }
    return L;
}
ll ans,sum[maxn],tmp;
int main()
{
    int N,K;
    scanf("%d%d",&N,&K);
    rep(i,1,N) scanf("%lld%lld",&a[i].x,&a[i].y);
    sort(a+1,a+K+1,cmp); convexhull(K);
    rep(i,1,top-1) ans+=det(ch[i],ch[i+1]),sum[i+1]=ans;
    rep(i,K+1,N){
        if(a[i].x<ch[1].x){
            int L=get(1,ttop,i,1),R=get(ttop,top,i,-1);
            tmp=sum[R]-sum[L]+det(ch[R],a[i])+det(a[i],ch[L]);
        }
        else if(a[i].x>ch[ttop].x){
            int L=get(1,ttop,i,-1),R=get(ttop,top,i,1);
            tmp=sum[top]-sum[R]+sum[L]+det(ch[L],a[i])+det(a[i],ch[R]);
        }
        else if(det(ch[ttop]-a[1],a[i]-ch[1])>0){//shang
            int Mid=bord(ttop,top,i,-1);
            if(Mid>ttop&&det(ch[Mid]-ch[Mid-1],a[i]-ch[Mid-1])>0) continue;
            int L=Mid>ttop?get(ttop,Mid-1,i,-1):Mid;
            int R=get(Mid,top,i,1);
            tmp=sum[top]-sum[R]+sum[L]+det(ch[L],a[i])+det(a[i],ch[R]);
        }
        else {
            int Mid=bord(1,ttop,i,1);
            if(Mid>1&&det(ch[Mid]-ch[Mid-1],a[i]-ch[Mid-1])>0) continue;
            int L=Mid>1?get(1,Mid-1,i,-1):1;
            int R=get(Mid,ttop,i,1);
            tmp=sum[top]-sum[R]+sum[L]+det(ch[L],a[i])+det(a[i],ch[R]);
        }
        ans=max(ans,tmp);
    }
    printf("%lld.%lld
",ans/2,ans%2*5);
    return 0;
}
/*
5 3
-5 -5
-5 5
5 -5
-4 6
5 5
*/