Codeforces Beta Round #64D - Professor's task

题意:两种操作1.加点2.查询点是否在之前给定点的凸包内
题解:set维护动态凸包,分别维护上下凸壳,对y取反就行,判断点是否在凸壳内,把点加进去看要不要删除就好了

//#pragma GCC optimize(2)
//#pragma GCC optimize(3)
//#pragma GCC optimize(4)
//#pragma GCC optimize("unroll-loops")
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include<bits/stdc++.h>
#define fi first
#define se second
#define db double
#define mp make_pair
#define pb push_back
#define pi acos(-1.0)
#define ll long long
#define vi vector<int>
#define mod 1000000007
#define ld long double
//#define C 0.5772156649
//#define ls l,m,rt<<1
//#define rs m+1,r,rt<<1|1
#define pll pair<ll,ll>
#define pil pair<int,ll>
#define pli pair<ll,int>
#define pii pair<int,int>
#define ull unsigned long long
//#define base 1000000000000000000
#define fin freopen("a.txt","r",stdin)
#define fout freopen("a.txt","w",stdout)
#define fio ios::sync_with_stdio(false);cin.tie(0)
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}
inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}
template<typename T>inline T const& MAX(T const &a,T const &b){return a>b?a:b;}
template<typename T>inline T const& MIN(T const &a,T const &b){return a<b?a:b;}
inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}
inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=ans*a%c;a=a*a%c,b>>=1;}return ans;}

using namespace std;

const ull ba=233;
const db eps=1e-8;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int N=200000+10,maxn=1000000+10,inf=0x3f3f3f3f;

struct node{
    ll x,y;
    node(){}
    node(ll _x,ll _y){x=_x,y=_y;}
    node operator -(const node&rhs)const{
        return node(x-rhs.x,y-rhs.y);
    }
    bool operator <(const node&rhs)const{
        return x<rhs.x||(x==rhs.x&&y<rhs.y);
    }
};
struct hull:set<node>{
    bool bad(iterator it)
    {
        if(it==begin()||next(it)==end())return 0;
        return (it->y-prev(it)->y)*(next(it)->x-it->x)<=(next(it)->y-it->y)*(it->x-prev(it)->x);
    }
    void update(node x)
    {
        if(find(x)!=end())return ;
        iterator it=insert(x).fi;
        if(bad(it)){erase(it);return ;}
        while(it!=begin()&&bad(prev(it)))
        {
//            ans-=dis(*it,*prev(it));
            erase(prev(it));
        }
        while(next(it)!=end()&&bad(next(it)))
        {
//            ans-=dis(*it,*next(it));
            erase(next(it));
        }
    }
    bool check(node x)
    {
        if(find(x)!=end())return 1;
        iterator it = insert(x).fi;bool ok=bad(it);erase(x);
        return ok;
    }
}h,h1;
int main()
{
    int n;scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        node a;int op;
        scanf("%d%lld%lld",&op,&a.x,&a.y);
        if(op==1)h.update(a),a.y=-a.y,h1.update(a);
        else
        {
            bool ok=h.check(a);a.y=-a.y;
            ok&=h1.check(a);
            puts(ok?"YES":"NO");
        }
    }
    return 0;
}
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标签：node,task,const,64D,Professor,ans,return,ll,define
来源： https://www.cnblogs.com/acjiumeng/p/10658102.html