CF70D Professor's task（动态凸包）

Posted 2021-02-06 bztminamoto

题面

两种操作:

1 往点集S中添加一个点(x,y);

2 询问(x,y)是否在点集S的凸包中. 数据保证至少有一个2操作, 保证刚开始会给出三个1操作, 且这三个操作中的点不共线.

题解

动态凸包板子

本来是习惯直接搞整个凸包的，这里似乎得分别维护上下凸壳，然后用平衡树来加点

话说没啥好说的……看代码啥都懂了……

//minamoto
#include<bits/stdc++.h>
#define R register
#define fi first
#define se second
#define IT map<int,int>::iterator
#define pi pair<int,int>
#define ll long long
#define fp(i,a,b) for(R int i=a,I=b+1;i<I;++i)
#define fd(i,a,b) for(R int i=a,I=b-1;i>I;--i)
#define go(head,u) for(int i=head[u],v=e[i].v;i;i=e[i].nx,v=e[i].v)
template<class T>inline bool cmax(T&a,const T&b){return a<b?a=b,1:0;}
using namespace std;
char buf[1<<21],*p1=buf,*p2=buf;
inline char getc(){return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++;}
int read(){
    R int res,f=1;R char ch;
    while((ch=getc())>'9'||ch<'0')(ch=='-')&&(f=-1);
    for(res=ch-'0';(ch=getc())>='0'&&ch<='9';res=res*10+ch-'0');
    return res*f;
}
map<int,int>up,dw;
int q,x,y,op;
inline ll cross(pi a,pi b,pi c){return 1ll*(b.fi-a.fi)*(c.se-a.se)-1ll*(b.se-a.se)*(c.fi-a.fi);}
bool ck(map<int,int> &mp,int x,int y){
    if(mp.empty())return false;
    if(mp.find(x)!=mp.end())return y>=mp[x];
    if(x<mp.begin()->fi||x>(--mp.end())->fi)return false;
    IT p=mp.lower_bound(x),q=p;--q;
    return cross(pi(x,y),*q,*p)>=0;
}
void ins(map<int,int> &mp,int x,int y){
    if(ck(mp,x,y))return;
    mp[x]=y;
    IT it=mp.upper_bound(x),itl=it;
    if(it!=mp.end()){
        ++itl;
        while(itl!=mp.end()&&cross(pi(x,y),*itl,*it)>=0)mp.erase(it),it=itl,++itl;
    }
    it=mp.lower_bound(x);IT itr=it;--itr;
    if(it==mp.begin()||itr==mp.begin())return;
    --it,--itr;
    while(it!=mp.begin()&&cross(pi(x,y),*it,*itr)>=0)mp.erase(it),it=itr,--itr;
}
int main(){
//  freopen("testdata.in","r",stdin);
    q=read();
    while(q--){
        op=read(),x=read(),y=read();
        if(op==1)ins(dw,x,y),ins(up,x,-y);
        else{
            bool fl1=ck(dw,x,y),fl2=ck(up,x,-y);
            puts((fl1&&fl2)?"YES":"NO");
        }
    }
    return 0;
}