Loj 2008 小凸想跑步

Loj 2008 小凸想跑步

• \(S(P,p_0,p_1)<S(P,p_i,p_{i+1})\) 這個約束條件對於 \(P_x,P_y\) 是線性的,即將面積用向量叉積表示,暴力拆開,可獲得 \(aP_x+bP_y+c<0\) 的形式,表示了一個半平面,其餘每條邊都肯定了一個半平面.
• 再將 \(P\) 在多邊形內拆成 \(N-1\) 個半平面的限制,將這 \(2N-1\) 個半平面求交,獲得的區域即爲合法區域,除以總面積即得答案

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
inline int read()
{
    int out=0,fh=1;
    char jp=getchar();
    while ((jp>'9'||jp<'0')&&jp!='-')
        jp=getchar();
    if (jp=='-')
        fh=-1,jp=getchar();
    while (jp>='0'&&jp<='9')
        out=out*10+jp-'0',jp=getchar();
    return out*fh;
}
const double eps=1e-8;
inline int dcmp(double x)
{
    if(fabs(x)<=eps)
        return 0;
    return x>0;
}
const int MAXN=2e5+10;
struct v2
{
    double x,y;
    v2(double x=0,double y=0):x(x),y(y) {}
    friend double operator * (const v2 &a,const v2 &b)
    {
        return a.x*b.y-a.y*b.x;
    }
    v2 operator + (const v2 &rhs) const
    {
        return v2(x+rhs.x,y+rhs.y);
    }
    v2 operator - (const v2 &rhs) const
    {
        return v2(x-rhs.x,y-rhs.y);
    }
    v2 operator ^ (const double &lambda) const
    {
        return v2(x*lambda,y*lambda);
    }
    double modulus()
    {
        return sqrt(x*x+y*y);
    }
    double angle()
    {
        return atan2(y,x);
    }
    bool operator < (const v2 &rhs) const
    {
        return x==rhs.x?y<rhs.y:x<rhs.x;
    }
};
struct Line
{
    v2 p,v;
    double angle()
    {
        return v.angle();
    }
    friend bool operator < (Line a,Line b)
    {
        if(a.angle()!=b.angle())
            return a.angle()<b.angle();
        return a.v*b.v<0;
    }
};
bool onleft(Line L,v2 p)
{
    return (L.v*(p-L.p))>0;
}
v2 intersection(Line a,Line b)
{
    v2 u=a.p-b.p;
    double t=(b.v*u)/(a.v*b.v);
    return a.p+(a.v^t);
}
#define x(I) poly[I].x
#define y(I) poly[I].y
int n,totl=0;
v2 poly[MAXN];
Line L[MAXN];
Line q[MAXN];
v2 p[MAXN];
int head,tail;
void Hpi()
{
    q[head=tail=1]=L[i];
    for(int i=2;i<=totl;++i)
    {
        while(head<tail && !Onleft(L[i],p[tail-1]))
            --tail;
        while(head<tail && !Onleft(L[i],p[head]))
            ++head;
        q[++tail]=L[i];
        if(head<tail && fabs(q[tail]*q[tail-1])==0)
        {
            --tail;
            if(Onleft(L[i].p,q[tail]))
                q[tail]=L[i];
        }
        if(head<tail)
            p[tail-1]=Intersection(q[tail-1],q[tail]);
    }
    while(head<tail && !Onleft(q[head],p[tail-1]))
        --tail;
    p[tail]=Intersection(p[tail],p[head]);
    p[tail+1]=p[1];
    double area=0;
    v2 O=v2(0,0);
    for(int i=head;i<=tail;++i)
        area+=fabs((O-p[i])*(O-p[i+1]));
    double ts=0;
    for(int i=1;i<=n;++i)
        ts+=fabs((O-poly[i])*(O-poly[i+1]));
    printf("%.4lf\n",area/ts);
}
int main()
{
    n=read();
    for(int i=1; i<=n; ++i)
        scanf("%lf%lf",&poly[i].x,&poly[i].y);
    poly[n+1]=poly[1];
    for(int i=1; i<=n; ++i)
    {
        ++totl;
        L[totl].p=poly[i];
        L[totl].v=poly[i+1]-poly[i];
    }
    for(int i=2; i<=n; ++i)
    {
        double a=y(1)-y(2)-y(i)+y(i+1);
        double b=x(2)-x(1)-x(i+1)+x(i);
        double c=(x(1)-x(1+1))*y(1)+(y(1+1)-y(1))*x(1);
        c-=(x(i)-x(i+1))*y(i)+(y(i+1)-y(i))*x(i);
        if(!b)
        {
            v2 p1=v2(-c/a,1.0);
            v2 p2=v2(-c/a,2.0);
            if(a>0)
                swap(p1,p2);
            ++totl;
            L[totl].p=p2;
            L[totl].v=p1-p2;
            continue;
        } 
        v2 p1=v2(0,-c/b);
        v2 p2=v2(1.0,-(a+c)/b);
        if(b<0)
            swap(p1,p2);
        ++totl;
        L[totl].p=p2;
        L[totl].v=p1-p2;
    }
    Hpi();
    return 0;
}