SWERC2017 Blowing Candles 凸包+旋转卡壳（模板题）

最新推荐文章于 2021-10-06 15:48:49 发布

原创最新推荐文章于 2021-10-06 15:48:49 发布 · 325 阅读

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#数学 #几何

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本文介绍了一种通过计算凸包并使用旋转卡尺算法来确定生日蛋糕上蜡烛所形成的两条平行线间最短距离的方法。该方法利用了C++编程语言中的几何运算和数据结构，适用于解决类似的几何问题。

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生日蛋糕上有蜡烛，问把蜡烛包括的两条平行线间的最短距离；

#include <iostream>
#include <string.h>
#include <algorithm> 
#include <math.h>
#include <vector>
using namespace std;
const int maxn=2e5+10;
struct point{
	double x, y;
    point(){}
    point(double x, double y) : x(x), y(y) {}
    friend point operator + (const point &a, const point &b){
        return point(a.x + b.x, a.y + b.y);
    }
    friend point operator - (const point &a, const point &b){
        return point(a.x - b.x, a.y - b.y);
    }
    friend point operator * (const point &a, const double &b){
        return point(a.x * b, a.y * b);
    }
    friend point operator / (const point &a, const double &b){
        return point(a.x / b, a.y / b);
    }
    friend bool operator == (const point &a, const point &b){
        return a.x == b.x && a.y == b.y;
    }
};
int r;
double dis(point a,point b)
{
	return sqrt((a.x-b.x)*(a.x-b.x)*1.0+(a.y-b.y)*(a.y-b.y));
}
double cross(point a,point b,point c) //计算叉乘 
{	
	return (c.x - a.x) * (b.y - a.y) - (b.x - a.x) * (c.y - a.y);	
} 
double  det(point a, point b)
{
    return a.x * b.y - a.y * b.x;
}
double dot(point a, point b)
{
    return a.x * b.x + a.y * b.y;
}
struct polygon_convex
{
    vector<point> p;
    polygon_convex(int size = 0){
        p.resize(size);
    }
};
bool comp_less(const point &a, const point &b)
{
    return a.x - b.x < 0 || a.x - b.x == 0 && a.y - b.y < 0;
}
polygon_convex convex_hull(vector<point> a)
{
    polygon_convex res(2 * a.size() + 5);
    sort(a.begin(), a.end(), comp_less);
    a.erase(unique(a.begin(), a.end()), a.end());
    int m = 0;
    for(int i = 0; i < a.size(); i ++){
        while(m > 1 && det(res.p[m - 1] - res.p[m - 2], a[i] - res.p[m - 2]) <= 0) -- m;
        res.p[m ++] = a[i];
    }
    int k = m;
    for(int i = int(a.size()) - 2; i >= 0; i --){
        while(m > k && det(res.p[m - 1] - res.p[m - 2], a[i] - res.p[m - 2]) <= 0) -- m;
        res.p[m ++] = a[i];
    }
    res.p.resize(m);
    if(a.size() > 1) res.p.resize(m - 1);
    return res;
}
double rotating_caliper(point *p,int n)
{
	 int j=1;
	 double minn=2.0*r;
	 p[n]=p[0];
	 for (int i=0;i<n;i++)
	 {
		 while (cross(p[i],p[i+1],p[j+1])-cross(p[i],p[i+1],p[j])<=0)
		 	j=(j+1)%n;
		//cout<<dis(p[i],p[i+1])<<"++++"<<endl;
		minn=min(minn,fabs(cross(p[i],p[i+1],p[j]))/dis(p[i],p[i+1]));
		//cout<<minn<<"^^^^^"<<endl;
	}
	return minn;
}
polygon_convex a;
point t,p[maxn];
int main()
{
	int nu,n;
	double s=0.0;
	scanf("%d%d",&nu,&r);
	for (int i=0;i<nu;i++)
	{
		scanf("%lf%lf",&t.x,&t.y);
		a.p.push_back(t);
	}
	a=convex_hull(a.p);
	n=a.p.size();
	for (int i=0;i<n;i++)
		p[i]=a.p[i];
//	for (int i=0;i<n;i++)
//		cout<<p[i].x<<"$$$$"<<p[i].y<<endl;
	if (n<=2)
		printf("0.00000000\n");
	else 
		printf("%.12lf\n",rotating_caliper(p,n));
	return 0;
}