#include<bits/stdc++.h>
#define eps 1e-9
#define ll long long 
#define sit set<node> :: iterator
using namespace std;

struct node{
	int x,y;
	friend node operator +(const node &x,const node &y){
		return (node){x.x+y.x,x.y+y.y};
	}
	friend node operator -(const node &x,const node &y){
		return (node){x.x-y.x,x.y-y.y};
	}
	friend ll operator ^(const node &x,const node &y){
		return x.x*y.y-y.x*x.y;
	}
	friend bool operator ==(const node &x,const node &y){
		return fabs(x.x-y.x)<=eps&&fabs(x.y-y.y)<=eps;
	}
}o;
ll dis(node x,node y){
	return (x.x-y.x)*(x.x-y.x)+(x.y-y.y)*(x.y-y.y);
}
bool operator <(node a,node b){
	ll tmp=a^b;
	if(tmp>0) return 1;
	else if(tmp==0&&dis(a,o)<dis(b,o)) return 1;
	return 0; 
}
set<node> s;
sit pre(sit it){
	if(it==s.begin()) it=s.end();
	return --it;
}
sit nxt(sit it){
	++it;
	if(it==s.end()) it=s.begin();
	return it;
}
bool query(node x){
	sit it=s.lower_bound(x);
	if(it==s.end()) it=s.begin();
	ll tmp=(x-*pre(it))^(*it-*pre(it));
	return tmp<=0;
}
void ins(node x){
	if(query(x)) return ;
	s.insert(x);
	sit it=nxt(s.find(x));
	while(s.size()>3&&((x- *(nxt(it)))^(*(it)- *(nxt(it))))<=0) s.erase(it),it=nxt(s.find(x));
	it=pre(s.find(x));
	while(s.size()>3&&((x-*(it))^(*(it)- *(pre(it))))>=0) s.erase(it);it=pre(s.find(x));
}
int main(){
	int q;
	scanf("%d",&q);
	o=(node){0,0};
	for(int i=1;i<=3;i++){
		int op;int x,y;
		scanf("%d%d%d",&op,&x,&y);
		s.insert((node){x,y});
	}
	q-=3;
	while(q--){
		int op;int x,y;
		scanf("%d%d%d",&op,&x,&y);
		if(op==1) ins((node){x,y});
		if(op==2){
			if(query((node){x,y})) puts("YES");
			else puts("NO");
		}
	}
	return 0;
}

以上是我写的一个动态凸包的代码，在样例

5
1 -1 0
1 1 0
1 0 2
2 0 1
2 0 3

将 $(1,0)$ 和 $(-1,0)$ 这两个点当做相同点去掉了，证据在于使用 $\text{multiset}$ 问题就解决了，请问这是什么情况