Loj 2008 小凸想跑步
- \(S(P,p_0,p_1)<S(P,p_i,p_{i+1})\) 这个约束条件对于 \(P_x,P_y\) 是线性的,即将面积用向量叉积表示,暴力拆开,可得到 \(aP_x+bP_y+c<0\) 的形式,表示了一个半平面,其他每条边都确定了一个半平面.
- 再将 \(P\) 在多边形内拆成 \(N-1\) 个半平面的限制,将这 \(2N-1\) 个半平面求交,得到的区域即为合法区域,除以总面积即得答案
#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; }.