「PKUSC2018」PKUSC

传送门

Solution 

考虑求每个点的贡献

等价于一个以OA长为半径的圆心为原点的圆在多边形内的弧对应的角度/\(2\pi\)

求弧度可以利用三角剖分

在原点的点要特判，采用射线法就可以了

Code 

#include <bits/stdc++.h>
#define reg register
#define ll long long
#define db double

using namespace std;

int read()
{
    int x=0,f=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch<='9'&&ch>='0'){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
	return x*f;
}

const db eps=1e-6,Pi=acos(-1.);
const int N=1005;
int sign(db x){return ((x>eps)-(x<-eps));}

struct P{
    db x,y;
    P(db x=0,db y=0):x(x),y(y){}
    P operator+(P o){return P(x+o.x,y+o.y);}
    P operator-(P o){return P(x-o.x,y-o.y);}
    P operator *(db o){return P(x*o,y*o);}
    P rev(){return P(-x,-y);}
    db operator *(P o){return x*o.x+y*o.y;}
    db operator ^(P o){return x*o.y-y*o.x;}
    db mo(){return sqrt(x*x+y*y);}
    db thi(P o){return acos(((*this)*o)/(mo()*o.mo()));}
}s[N],p[N];

db calc(P a,P b,db R)
{
    db ras=0;
    db f=sign(a^b);
    if(sign(max(a.mo(),b.mo())-R)==-1) return 0;

    db thi=a.thi(b);
    db h=fabs((a^b)/(a-b).mo());
    if(sign(h-R)>-1) return thi*f;

    db a0=asin(h/R),a1=a.rev().thi(b-a),a2=b.rev().thi(a-b);
    if(sign(a0-a1)==1) ras+=a0-a1;
    if(sign(a0-a2)==1) ras+=a0-a2;
    return min(ras,thi)*f;
}

int chk(P x,P y)
{
    if(!sign(x.y)) return sign(x.x)==1;
    if(!sign(y.y)) return 0;
    if(x.y>y.y) swap(x,y);
    if(sign(x.y*y.y)==1) return 0;
    P xxx=x+(y-x)*(x.y/(y.y-x.y));
    return sign(xxx.x)==1;
}

int main()
{
    reg int i,j,n,m;
    db ans=0.;
    n=read(),m=read();

    for(i=1;i<=n;++i) p[i].x=read(),p[i].y=read();
    for(i=1;i<=m;++i) s[i].x=read(),s[i].y=read();
    s[m+1]=s[1];

    for(i=1;i<=m;++i)
    {
        if(!sign(s[i]^s[i+1])) continue;
        for(j=1;j<=n;++j)
        {
            if(!sign(p[j].mo())) continue;
            ans+=calc(s[i],s[i+1],p[j].mo());
        }
    }

    for(i=1;i<=n;++i)
	{
        if(sign(p[i].mo())) continue;
        int flag=1,cnt=0;
        for(j=1;j<=m;++j)
        {
            if(sign(s[j]^s[j+1])) cnt+=chk(s[j],s[j+1]);
            else if(sign(s[j]*s[j+1])<1){flag=0;break;}
        }
        if(flag&&cnt%2==1) ans+=Pi*2.;
    }

    ans/=(Pi*2.);

    return 0*printf("%.5lf\n",ans);
}

---

Blog来自PaperCloud，未经允许，请勿转载，TKS！