Description

• 题目给你一个凸多边形，可以绕原点随便转，剪刀固定方向，限制长度，求可以剪开的概率

Solution

• 这是一种巧妙的暴力做法
• 转多边形很麻烦，我们相对地想到转剪刀方向
• 我们每次把剪刀转一点点
• “一点点”来自精度要求$10^{-4}$，$\frac{2\pi }{3\ast 10^5}rad$绰绰有余
• 算出剪刀沿此方向需要剪的距离，与$L$比较
• 如果超过$L$说明剪不开

#include <bits/stdc++.h>
const int N=1e5+10;
const double P=acos(0.0)*2.0,eps=1e-8;
using namespace std;
int n;double d,L;
int sgn(double x){
    return (fabs(x)<eps)?0:(x<0?-1:1);}
struct Point{
    
	double x,y;
	Point(){
    }
	Point(double _x,double _y){
    x=_x,y=_y;}
	Point operator +(Point a){
    return Point(x+a.x,y+a.y);}//向量加
	Point operator -(Point a){
    return Point(x-a.x,y-a.y);}//减
	double operator *(Point a){
    return x*a.x+y*a.y;}//点积
	Point operator %(double a){
    return Point(x*a,y*a);}//数乘（伸缩）Scaling
	double operator ^(Point a){
    return x*a.y-y*a.x;}//叉积
}v[N<<2];
struct Line{
    Point s,e;Line(){
    }Line(Point _s,Point _e){
    s=_s,e=_e;}}ln;
Point rotate(Point c,double a){
    return Point(c.x*cos(a)-c.y*sin(a),c.x*sin(a)+c.y*cos(a));}
//点绕原点转a rad
Point LIL(Line a,Line b){
    return a.s+(a.e-a.s)%(((b.e-b.s)^(b.s-a.s))/((b.e-b.s)^(a.e-a.s)));}
//Line Intersects Line,直线交直线
double displ(Point a,Line b){
    
	Point v1=b.e-b.s,v2=a-b.s;
	return fabs(v1^v2)/sqrt(v1.x*v1.x+v1.y*v1.y);
}int main(){
    
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
		scanf("%lf%lf",&v[i].x,&v[i].y),v[i+n]=v[i],v[i+2*n]=v[i],v[i+3*n]=v[i];
	scanf("%lf%lf%lf%lf%lf",&L,&ln.s.x,&ln.s.y,&ln.e.x,&ln.e.y);	
	d=displ(Point(0,0),ln);//剪刀到原点的距离
	Point p0(0.0,0.0);
	int cnt=0,tot=3e5,l=1,r=1;
	for(int i=0;i<tot;++i){
    
		double a=i*2*P/tot;
		Point cur=Point(cos(a),sin(a))%d;//单位圆定向,d定长
		Point vv=rotate(cur,P/2);//剪刀与距离成90角
		Point nxt=cur+vv;//中间量
		Point p1,p2;
		Line s0=Line(cur,nxt);//真正的剪刀方向
		while(true){
    
			int o1=sgn((nxt-cur)^(v[l]-nxt));
			int o2=sgn((nxt-cur)^(v[l+1]-nxt));
			if(o1*o2==1){
    ++l;continue;}
			p1=LIL(s0,Line(v[l],v[l+1]));//往一边找到与多边形相交的点
			break;
		}r=max(r,l+1);
		while(true){
    
			int o1=sgn((nxt-cur)^(v[r]-nxt));
			int o2=sgn((nxt-cur)^(v[r+1]-nxt));
			if(o1*o2==1){
    ++r;continue;}
			p2=LIL(s0,Line(v[r],v[r+1]));//往另一边找到与多边形相交的点
			break;
		}Point p=p1-p2;//剪刀剪开需要剪过的向量
		double res=sqrt(p.x*p.x+p.y*p.y);//|p|
		if(sgn(L-res)<0) ++cnt;
	}printf("%.4lf\n",1.0*cnt/tot);
}