计算几何+求质心+求多边形与圆交面积——ICPC GNYR 2019

/*
先求出多边形的质心
    将多边形分割成三角形 OViVi+1,
    求每个三角形的质心((x1+x2+x3)/3,(y1+y2+y3)/3)，然后再有向面积加权 
然后求圆和多边形的交点：套模板即可 
*/
#include<bits/stdc++.h>
using namespace std;

typedef double db;
const db eps=1e-6;
const db pi=acos(-1);
int sign(db k){
    if (k>eps) return 1; else if (k<-eps) return -1; return 0;
}
int cmp(db k1,db k2){return sign(k1-k2);}
int inmid(db k1,db k2,db k3){return sign(k1-k3)*sign(k2-k3)<=0;}// k3 在 [k1,k2] 内 
struct point{
    db x,y;
    point operator + (const point &k1) const{return (point){k1.x+x,k1.y+y};}
    point operator - (const point &k1) const{return (point){x-k1.x,y-k1.y};}
    point operator * (db k1) const{return (point){x*k1,y*k1};}
    point operator / (db k1) const{return (point){x/k1,y/k1};}
    int operator == (const point &k1) const{return cmp(x,k1.x)==0&&cmp(y,k1.y)==0;}
    point turn(db k1){return (point){x*cos(k1)-y*sin(k1),x*sin(k1)+y*cos(k1)};}
    point turn90(){return (point){-y,x};}
    bool operator < (const point k1) const{
        int a=cmp(x,k1.x);
        if (a==-1) return 1; else if (a==1) return 0; else return cmp(y,k1.y)==-1;
    }
    db abs(){return sqrt(x*x+y*y);}
    db abs2(){return x*x+y*y;}
    db dis(point k1){return ((*this)-k1).abs();}
    point unit(){db w=abs(); return (point){x/w,y/w};}
};
int inmid(point k1,point k2,point k3){return inmid(k1.x,k2.x,k3.x)&&inmid(k1.y,k2.y,k3.y);}
db cross(point k1,point k2){return k1.x*k2.y-k1.y*k2.x;}
db dot(point k1,point k2){return k1.x*k2.x+k1.y*k2.y;}
db rad(point k1,point k2){return atan2(cross(k1,k2),dot(k1,k2));}
point proj(point k1,point k2,point q){ // q 到直线 k1,k2 的投影 
    point k=k2-k1; return k1+k*(dot(q-k1,k)/k.abs2());
}
db disSP(point k1,point k2,point q){
    point k3=proj(k1,k2,q);
    if (inmid(k1,k2,k3)) return q.dis(k3); else return min(q.dis(k1),q.dis(k2));
}
struct circle{
    point o; db r;
    int inside(point k){return cmp(r,o.dis(k));}
};
vector<point> getCL(circle k1,point k2,point k3){ // 求直线和圆的交点，沿着 k2->k3 方向给出 , 相切给出两个 
    point k=proj(k2,k3,k1.o); db d=k1.r*k1.r-(k-k1.o).abs2();
    if (sign(d)==-1) return {};
    point del=(k3-k2).unit()*sqrt(max((db)0.0,d)); return {k-del,k+del};
}
db getarea(circle k1,point k2,point k3){
    // 圆 k1 与三角形 k2 k3 k1.o 的有向面积交
    point k=k1.o; k1.o=k1.o-k; k2=k2-k; k3=k3-k;
    int pd1=k1.inside(k2),pd2=k1.inside(k3); 
    vector<point>A=getCL(k1,k2,k3);
    if (pd1>=0){
        if (pd2>=0) return cross(k2,k3)/2;
        return k1.r*k1.r*rad(A[1],k3)/2+cross(k2,A[1])/2;
    } else if (pd2>=0){ 
        return k1.r*k1.r*rad(k2,A[0])/2+cross(A[0],k3)/2;
    }else {
        int pd=cmp(k1.r,disSP(k2,k3,k1.o));
        if (pd<=0) return k1.r*k1.r*rad(k2,k3)/2;
        return cross(A[0],A[1])/2+k1.r*k1.r*(rad(k2,A[0])+rad(A[1],k3))/2;
    }
}


point p[200],k;
circle c;
db X,Y,sum;
int n;

int main(){
    cin>>n;
    for(int i=0;i<n;i++)cin>>p[i].x>>p[i].y;
    k.x=k.y=0;
    for(int i=0;i<n;i++)//求总面积 
        sum+=cross(p[i]-k,p[(i+1)%n]-k)/2;
    for(int i=0;i<n;i++){
        db area=cross(p[i]-k,p[(i+1)%n]-k)/2;
        db x=(p[i].x+p[(i+1)%n].x)/3;
        db y=(p[i].y+p[(i+1)%n].y)/3;
        X+=x*area/sum;
        Y+=y*area/sum;
    }
    c.o=(point){X,Y};
    c.r=sqrt(sum/pi); //sum=pi*r*r
    
    db ans=0; 
    for(int i=0;i<n;i++){ //求相交面积 
        ans+=getarea(c,p[i],p[(i+1)%n]);
    }
    
    printf("%.4lf\n",ans/sum);
}