【BZOJ4445】[SCOI2015]小凸想跑步（半平面交）

题面

BZOJ

洛谷

题解

首先把点给设出来，\(A(x_a,y_a),B(x_b,y_b),C(x_c,y_c),D(x_d,y_d),P(x,y)\)

然后我们考虑\(S_\Delta ABP<S_\Delta CDP\)什么情况下满足。

根据点积来求面积，得到：

这个东西左边拆开之后得到：

右侧类似。然后就可以移项，得到：

那么对于相邻的两个点和\(0,1\)两个点进行一次比较，每次都可以得到一个半平面，最后求解这个半平面交就是结果了。

#include<iostream>

#include<cstdio>

#include<cmath>

#include<algorithm>

using namespace std;

#define MAX 100100

const double eps=1e-7;

inline int read()

{

	int x=0;bool t=false;char ch=getchar();

	while((ch<'0'||ch>'9')&&ch!='-')ch=getchar();

	if(ch=='-')t=true,ch=getchar();

	while(ch<='9'&&ch>='0')x=x*10+ch-48,ch=getchar();

	return t?-x:x;

}

struct Node{double x,y;}p[MAX],Qp[MAX<<1];

Node operator+(Node a,Node b){return (Node){a.x+b.x,a.y+b.y};}

Node operator-(Node a,Node b){return (Node){a.x-b.x,a.y-b.y};}

Node operator*(Node a,double b){return (Node){a.x*b,a.y*b};}

double operator*(Node a,Node b){return a.x*b.x+a.y*b.y;}

double cross(Node a,Node b){return a.x*b.y-a.y*b.x;}

struct Line{Node p,v;double alpha;}S[MAX<<1],Q[MAX<<1];int tot;

bool cmp(Line a,Line b){return a.alpha<b.alpha;}

Node Intersection(Line a,Line b)

{

	Node c=b.p-a.p;

	double t=cross(b.v,c)/cross(b.v,a.v);

	return a.p+a.v*t;

}

int zero(double x)

{

	if(fabs(x)<eps)return 0;

	return x>0?1:-1;

}

bool Left(Node a,Line b){return zero(cross(b.v,a-b.p))>0;}

int n;double area;

int l,r;

bool HalfPlaneIntersection()

{

	for(int i=1;i<=tot;++i)S[i].alpha=atan2(S[i].v.y,S[i].v.x);

	sort(&S[1],&S[tot+1],cmp);

	Q[l=r=1]=S[1];

	for(int i=2;i<=tot;++i)

	{

		while(l<r&&!Left(Qp[r-1],S[i]))--r;

		while(l<r&&!Left(Qp[l],S[i]))++l;

		if(zero(cross(Q[r].v,S[i].v))==0)

			Q[r]=Left(Q[r].p,S[i])?Q[r]:S[i];

		else Q[++r]=S[i];

		if(l<r)Qp[r-1]=Intersection(Q[r],Q[r-1]);

	}

	while(l<r&&!Left(Qp[r-1],Q[l]))--r;

	return (r-l)>1;

}

int main()

{

	n=read();

	for(int i=0;i<n;++i)p[i].x=read(),p[i].y=read();p[n]=p[0];

	for(int i=0;i<n;++i)S[++tot]=(Line){p[i],p[i+1]-p[i]};

	for(int i=1;i<n;++i)area+=fabs(cross(p[i]-p[0],p[i+1]-p[0]));

	for(int i=1;i<n;++i)

	{

		double a=p[0].y-p[1].y-p[i].y+p[i+1].y;

		double b=-p[0].x+p[1].x+p[i].x-p[i+1].x;

		double c=cross(p[0],p[1])-cross(p[i],p[i+1]);

		Line d;

		if(fabs(a)>eps)d=(Line){(Node){-c/a,0},(Node){-b,a}};

		else d=(Line){(Node){0,-c/b},(Node){-b,a}};

		S[++tot]=d;

	}

	HalfPlaneIntersection();

	double ans=0;Qp[r]=Intersection(Q[l],Q[r]);

	for(int i=l+1;i<r;++i)ans+=fabs(cross(Qp[i]-Qp[l],Qp[i+1]-Qp[l]));

	printf("%.4lf\n",ans/area);

	return 0;

}