BIG # 6437. 「PKUSC2018」 PKUSC

Face questions

Meaning into question:

How long circular arc is determined at each point where the corner is inside the polygon.

And then very violent.

Each point P, directly to the polygon and the circle of intersection of all, the angle P is sorted according to distance.

Find intersection, direct simultaneous two yuan quadratic equations. . . .

 

We need to determine whether an arc inside the polygon.

Random rotation angle of the vector, to determine whether the point within the polygon.

If the polygon edge point, it returns -1, spinning again.

Due to double, in the case of the edge of a polygon-ray does not occur.

 

note:

(0,0) for a special judge whether inside the polygon. Analyzing + eps

#include<bits/stdc++.h>
#define reg register int
using namespace std;
// using namespace Modulo;
namespace Miracle{
const int N=505;
const double eps=1e-8;
const double inf=1e8;
const double Pi=acos(-1);
int n,m;
struct po{
    double x,y;
    po () {}
    po(double xx,double yy){
        x=xx;y=yy;
    }
    double friend operator *(po a,po b){
        return a.x*b.y-a.y*b.x;
    }
    double dis(){
        return sqrt(x*x+y*y);
    }
    po friend operator -(po a,po b){
        return po(a.x-b.x,a.y-b.y);
    }
    double deg(){
        return atan2(y,x);
    }
    po xuan(double d){
        double nd=deg()-d;
        double len=dis();
        return po(len*cos(nd),len*sin(nd));
    }
    void up () {
        cout<<"("<<x<<","<<y<<")"<<endl;
    }
}p[N],q[N];
struct node{
    when P;
    double c;
    node(){}
    node(double cc,po PP){
        c = cc; P = PP;
    }
    bool friend operator <(node a,node b){
        return a.c<b.c;
    }
}cur[N];
struct line{
    double k,b;
    po A, B;
    double f(double x){
        if(k<inf) return k*x+b;
        return A.y;
    }
}l[N];
bool on(po c,line L){//c on L?
    if(c.x+eps>=L.A.x&&c.x<=L.B.x+eps){
        if(L.k==inf) {
            if(c.y+eps>=L.A.y&&c.y<=L.B.y+eps) return 1;
        }
        else{double ny=L.f(c.x);
        if(fabs(ny-c.y)<eps) return 1;}
    }
    return 0;
}
bool jiao(line a,line b){
    double c1=(a.A-b.A)*(a.A-b.B),c2=(a.B-b.A)*(a.B-b.B);
    double c3=(b.A-a.A)*(b.A-a.B),c4=(b.B-a.A)*(b.B-a.B);
    if(c1*c2<=0&&c3*c4<=0) return true;
    return false;
}
int  in (Po) {
     int cnt = 0 ;
    line now;
    now.k=0;now.b=a.y;
    now.A=a;now.B=po(inf,a.y);
    for(reg i=1;i<=m;++i){
        if(on(a,l[i])) return -1;
        if(jiao(now,l[i])) ++cnt;
    }
    if(cnt&1) return 1;
    return 0;
}
double cha(double d1,double d2){
    if(d1>=0&&d2<=0){
        return d1-d2;
    }else if(d1>=0&&d2>=0){
        if(d1>d2) return d1-d2;
        return 2*Pi-(d2-d1);
    }else if(d1<=0&&d2>=0){
        return 2*Pi-(d2-d1);
    }else if(d1<=0&&d2<=0){
        if(d1>d2) return (d1-d2);
        return 2*Pi-(d2-d1);
    }
    return 0;
}
double sui(){
    return (double)rand()/(RAND_MAX);
}
double calc(po a){
    int cnt=0;
    double r=a.dis();
    if(r==0){
        a.y+=eps;
        if(in(a)==1) return 2*Pi;
        else return 0;
    }
    for(reg i=1;i<=m;++i){
        if(l[i].k<inf){
            double A=(1+l[i].k*l[i].k);
            double B=2*l[i].k*l[i].b;
            double C=l[i].b*l[i].b-r*r;
            double deta=B*B-4*A*C;
            if(deta>=0){
                double x1=(-B+sqrt(deta))/(2*A),x2=(-B-sqrt(deta))/(2*A);
                po now=po(x1,l[i].f(x1));
                if(on(now,l[i])) {
                    cur[++cnt]=node(cha(a.deg(),now.deg()),now);
                }
                now=po(x2,l[i].f(x2));
                if(on(now,l[i])){
                    cur[++cnt]=node(cha(a.deg(),now.deg()),now);
                }
            }
        }else{
            double lp=r*r-l[i].A.x*l[i].A.x;
            double X=l[i].A.x;
            if(lp>0){
                double Y=sqrt(lp);
                po now=po(X,Y);
                if(on(now,l[i])) {
                    cur[++cnt]=node(cha(a.deg(),now.deg()),now);
                }
                Y=-sqrt(lp);
                now=po(X,Y);
                if(on(now,l[i])){
                    cur[++cnt]=node(cha(a.deg(),now.deg()),now);
                }
            }
        }
    }
    
    cur[++cnt]=node(0,a);
    cur [ ++ cnt] = node ( 2 * Pi, a);
    sort(cur+1,cur+cnt+1);
    double ret=0;
    for(reg i=2;i<=cnt;++i){
        double d=cur[i].c-cur[i-1].c;
        if(d<0.0000001) continue;
        po now=cur[i-1].P;
        int tmp=-1;
        while(tmp==-1){
            double z=d*sui();
            now=cur[i-1].P.xuan(z);
            tmp=in(now);
        }
        if(tmp==1){
            entitled + = d;
        }
    }
    Return the right;

}
int main () {
    srand((unsigned long long)new char);
    cin>>n>>m;
    for(reg i=1;i<=n;++i){
        double x,y;
        scanf("%lf%lf",&x,&y);
        q[i]=po(x,y);
    }
    for(reg i=1;i<=m;++i){
        double x,y;
        scanf("%lf%lf",&x,&y);
        p[i]=po(x,y);
    }
    for(reg i=1;i<=m;++i){
        int nxt=(i==m)?1:i+1;
        l[i].A=p[i];l[i].B=p[nxt];
        if(p[i].x>p[nxt].x||(p[i].x==p[nxt].x&&p[i].y>p[nxt].y)) swap(l[i].A,l[i].B);

        if(p[i].x==p[nxt].x){
            l[i].k=inf;
        }else{
            l[i].k=(p[nxt].y-p[i].y)/(p[nxt].x-p[i].x);
            l[i].b=((p[nxt].x*p[i].y)-(p[i].x*p[nxt].y))/(p[nxt].x-p[i].x);
        }
    }

    double ans=0.0;
    for(reg i=1;i<=n;++i){
        years + = calc (q [i]);
    }
    years = years / ( 2 * pi);
    printf("%.5lf",ans);
    return 0;
}

}
signed main(){
    Miracle::main();
    return 0;
}

/*
   Author: *Miracle*
*/