题面

BZOJ

题解

我卡精度卡了一天。。。。
之前不强制在线的做法是$CDQ$分治，维护一个凸壳就好了。
现在改成二进制分组，每次重建凸壳就好了。。

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
using namespace std;
#define ll long long
#define MAX 500500
#define Sqr(x) ((x)*(x))
#define pb push_back
struct Node{double x,y;}a[MAX];
bool operator<(Node a,Node b){if(a.x==b.x)return a.y<b.y;return a.x<b.x;}
inline double Slope(Node a,Node b)
{
    if(a.x==b.x)return a.y>b.y?1e18:-1e18;
    return (a.y-b.y)/(a.x-b.x);
}
int n,ans,top;
struct Group
{
    vector<Node> p,Q;
    int tot,tp;
    void insert(Node x){++tot;p.pb(x);}
    void clear(){p.clear();Q.clear();tot=tp=0;}
    void Build()
        {
            sort(p.begin(),p.end());
            tp=1;Q.clear();Q.pb(p[0]);
            for(int i=1;i<tot;++i)
            {
                while(tp>1&&Slope(Q[tp-1],Q[tp-2])-Slope(Q[tp-1],p[i])>=0)--tp,Q.pop_back();
                Q.pb(p[i]),++tp;
            }
        }
    bool Query(double x,double y)
        {
            double k=-x/y;int l=1,r=tp-1,ret=0;
            while(l<=r)
            {
                int mid=(l+r)>>1;
                if(k>=Slope(Q[mid],Q[mid-1]))l=mid+1,ret=mid;
                else r=mid-1;
            }
            return 2*x*Q[ret].x+2*y*Q[ret].y>=x*x+y*y;
        }
}B[50];
void insert(double x,double y)
{
    B[++top].insert((Node){x,y});
    while(top>1&&B[top].tot==B[top-1].tot)
    {
        for(int i=0;i<B[top].tot;++i)
            B[top-1].insert(B[top].p[i]);
        B[top--].clear();
    }
    B[top].Build();
}
bool Query(double x,double y)
{
    if(!top)return false;
    for(int i=1;i<=top;++i)
        if(!B[i].Query(x,y))return false;
    return true;
}
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;++i)
    {
        int opt;double x,y;
        scanf("%d%lf%lf",&opt,&x,&y);
        x+=ans;y+=ans;
        if(!opt)insert(x,y);
        else if(Query(x,y))++ans,puts("Yes");else puts("No");
    }
    return 0;
}