计算几何,向量——cf995c

网上的题解直接用随机过的，

自己用模拟就模拟三个向量的和并就模拟不出来。。

以后再回头看看

#include<bits/stdc++.h>
#include<cmath>
using namespace std;
 
const double esp = 1e-8;
const int maxn = 1e5+10;
const double C = 180.0/acos(-1.0);
struct Vector{
    double x,y,len,angle;
    int k,id;
}p[maxn];
int cmp(Vector a,Vector b){return a.len>b.len;}
int n;
 
void reverse(Vector &v){
    v.k*=-1;v.x*=-1;v.y*=-1;
    v.angle=atan2(v.y,v.x)*C;
    if(v.angle<0)v.angle+=360;
}
void add(Vector &a, Vector &b){
    a.x+=b.x;a.y+=b.y;
    a.len=sqrt(a.x*a.x+a.y*a.y);
    a.angle=atan2(a.y,a.x)*C;
    if(a.angle<0)a.angle+=360;
}
 
void Merge(Vector &a,Vector &b,Vector &c){//把c向量加到a向量或者b向量里 
    double r1=a.angle,r2=c.angle;
    if(r1>r2)swap(r1,r2);
    double A=min(fabs(r2-r1),fabs(r1-r2+360));
    if(A<=180 && A>=120){//形成钝角 
        add(a,c);return; 
    }
    else if(A<=60){//形成锐角 
        reverse(c);
        add(a,c);
        return;
    }
    
    r1=b.angle,r2=c.angle;
    if(r1>r2)swap(r1,r2);
    double B=min(fabs(r2-r1),fabs(r1-r2+360));
    if(B<=180 && B>=120){
        add(b,c);return;
    }
    else if(B<=60){//形成锐角 
        reverse(c);
        add(b,c);
        return;
    }
    
    Vector tmp=b;b=c;c=tmp;
    
    r1=a.angle,r2=c.angle;
    if(r1>r2)swap(r1,r2);
    A=min(fabs(r2-r1),fabs(r1-r2+360));
    if(A<=180 && A>=120){//形成钝角 
        add(a,c);return; 
    }
    else if(A<=60){//形成锐角 
        reverse(c);
        add(a,c);
        return;
    }
}
 
int main(){
    cin>>n;
    for(int i=1;i<=n;i++){
        scanf("%lf%lf",&p[i].x,&p[i].y);
        p[i].len=sqrt(p[i].x*p[i].x+p[i].y*p[i].y);
        p[i].angle=atan2(p[i].y,p[i].x)*C;
        if(p[i].angle<0)
            p[i].angle+=360; 
        p[i].id=i;p[i].k=1;
    }
    sort(p+1,p+1+n,cmp);
    
    Vector &a=p[1], &b=p[2];
    for(int i=3;i<=n;i++){
        Vector &c=p[i];
        Merge(a,b,c);
        if((a.x+b.x)*(a.x+b.x)>=2.25*1e12)
            cout<<i<<" NO";
    
    }
    /*
cout<<a.x<<" "<<a.y<<'\n';
cout<<b.x<<" "<<b.y<<'\n';*/

    int ans[maxn]={};
    for(int i=1;i<=n;i++)
        ans[p[i].id]=p[i].k;
    for(int i=1;i<=n;i++)cout<<ans[i]<<" ";
}