(hdu step 7.1.5)Maple trees(凸包的最小半径寻找掩护轮)

称号：

Maple trees

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 177 Accepted Submission(s): 63

Problem Description

There are a lot of trees in HDU. Kiki want to surround all the trees with the minimal required length of the rope . As follow, 

To make this problem more simple, consider all the trees are circles in a plate. The diameter of all the trees are the same (the diameter of a tree is 1 unit). Kiki can calculate the minimal length of the rope , because it's so easy for this smart girl.
But we don't have a rope to surround the trees. Instead, we only have some circle rings of different radius. Now I want to know the minimal required radius of the circle ring. And I don't want to ask her this problem, because she is busy preparing for the examination.
As a smart ACMer, can you help me ?

 

Input

The input contains one or more data sets. At first line of each input data set is number of trees in this data set n （1 <= n <= 100）, it is followed by n coordinates of the trees. Each coordinate is a pair of integers, and each integer is in [-1000, 1000], it means the position of a tree’s center. Each pair is separated by blank.
Zero at line for number of trees terminates the input for your program.

 

Output

Minimal required radius of the circle ring I have to choose. The precision should be 10^-2.

 

Sample Input

2
1 0
-1 0
0

 

Sample Output

1.50

 

Author

zjt

 

Recommend

lcy

 

题目分析：

求凸包的最小覆盖圆的半径。事实上就是在求完凸包以后再求一下最小覆盖圆即可了。

这道题须要用到下面的一些知识：

1、关于钝角三角形，假设c是斜边，那么必定有a^2 + b^2 < c^2的证明。

2、由三角形的三个顶点求一个三角形的面积。

已知三角形△A1A2A3的顶点坐标Ai ( xi , yi ) ( i =1, 2, 3) 。该三角形的面积为:

　　S =  ( (x2 - x1) * (y3 - y1) - (x3 - x1) * (y2 - y1) ) / 2 ;

　　△A1A2A3 边界构成逆时针回路时取+ , 顺时针时取 -。

　　另外在求解的过程中。不须要考虑点的输入顺序是顺时针还是逆时针，相除后就抵消了。

3、

 凸包+最小圆覆盖
 枚举随意3点找其最小覆盖圆
（当为钝角三角形时不是外接圆，而是以其最长边为直径的圆）。
 当为外接圆时，半径公式为r=abc/4s;(推导为例如以下：
 由正弦定理，a/sinA=b/sinB=c/sinC=2R,得sinA=a/(2R),
 又三角形面积公式Ｓ=(bcsinA)/2,所以S=(abc)/(4R),故R=(abc)/(4S).

这道题还须要注意的是：

1、在使用完graham求最小凸包以后。尽量让这个凸包闭合。即p[n] = p[0]。

代码例如以下：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>

using namespace std;

const double epsi = 1e-8;
const double pi = acos(-1.0);
const int maxn = 101;

struct PPoint{//结构体尽量不要定义成Point这样的,容易和C/C++本身中的变量同名
	double x;
	double y;

	PPoint(double _x = 0,double _y = 0):x(_x),y(_y){

	}

	PPoint operator - (const PPoint& op2) const{
		return PPoint(x - op2.x,y - op2.y);
	}

	double operator^(const PPoint &op2)const{
		return x*op2.y - y*op2.x;
	}
};

inline int sign(const double &x){
	if(x > epsi){
		return 1;
	}

	if(x < -epsi){
		return -1;
	}

	return 0;
}

inline double sqr(const double &x){
	return  x*x;
}

inline double mul(const PPoint& p0,const PPoint& p1,const PPoint& p2){
	return (p1 - p0)^(p2 - p0);
}

inline double dis2(const PPoint &p0,const PPoint &p1){
	return sqr(p0.x - p1.x) + sqr(p0.y - p1.y);
}

inline double dis(const PPoint& p0,const PPoint& p1){
	return sqrt(dis2(p0,p1));
}

int n;
PPoint p[maxn];
PPoint convex_hull_p0;

inline bool convex_hull_cmp(const PPoint& a,const PPoint& b){
	return sign(mul(convex_hull_p0,a,b)>0)|| (sign(mul(convex_hull_p0,a,b)) == 0 && dis2(convex_hull_p0,a) < dis2(convex_hull_p0,b));
}

int convex_hull(PPoint* a,int n,PPoint* b){
	int i;
	for(i = 1 ; i < n ; ++i){
		if(sign(a[i].x - a[0].x) < 0 || (sign(a[i].x - a[0].x) == 0 && sign(a[i].y - a[0].y) < 0)){
			swap(a[i],a[0]);
		}
	}

	convex_hull_p0 = a[0];//这两行代码不要顺序调换了..否则会WA
	sort(a,a+n,convex_hull_cmp);

	b[0] = a[0];
	b[1] = a[1];
	int newn = 2;
	for(i = 2 ; i < n ; ++i){
		while(newn > 1 && sign(mul(b[newn-1],b[newn-2],a[i])) >= 0){
			newn--;
		}

		b[newn++] = a[i];
	}

	return newn;
}

/**
 * 有一个三角形的三个点来计算这个三角形的面积
 */
double crossProd(PPoint A, PPoint B, PPoint C) {
    return (B.x-A.x)*(C.y-A.y) - (B.y-A.y)*(C.x-A.x);
}

int main(){
	while(scanf("%d",&n)!=EOF,n){
		int i;
		for(i = 0 ; i < n ; ++i){
			scanf("%lf %lf",&p[i].x,&p[i].y);
		}

		/**
		 * 处理节点数仅仅有1、2的情况
		 */
		if(n == 1){
			printf("0.50\n");
			continue;
		}
		if(n == 2){
			printf("%.2lf\n",dis(p[0],p[1])/2 + 0.5);
			continue;
		}

		/**
		 * 当结点数>=3时,用graham算法来求最小凸包
		 */
		n = convex_hull(p,n,p);
		p[n] = p[0];//记得要收尾相接,否则可能会出错

		int j;
		int k;

		double maxr = -1;//用于求最小覆盖圆的半径
		double r;
		/**
		 * 枚举凸包中的随意三个点.
		 * 假设这三个点形成的外接圆的半径最大,
		 * 那么这个就是我们所要找的凸包的最小覆盖圆
		 */
		for(i = 0 ; i < n ; ++i){
			for(j = i+1 ; j < n ; ++j){
				for(k = j+1 ; k <= n ; ++k){//注意,这里的k是 <= n
					double a = dis(p[i],p[j]);
					double b = dis(p[i],p[k]);
					double c = dis(p[j],p[k]);

					//假设这三个点所形成的是钝角三角形
					if(a*a+b*b < c*c || a*a+c*c < b*b || b*b+c*c < a*a){
						r = max(max(a,b),c)/2;//那么这时候的半径等于最长边的一半
					}else{//假设是直角三角形||锐角三角形
						double s = fabs(crossProd(p[i],p[j],p[k]))/2;//由定理1求得面积
						r = a*b*c/(4*s);//三角形的外接圆公式
					}

					if(maxr < r){//假设眼下存储的最大半径<当前外接圆的半径
						maxr = r;//则更新眼下的最大半径
					}
				}
			}
		}
		printf("%.2lf\n",maxr + 0.5);//输出凸包的最小覆盖圆的最大半径
	}

	return 0;
}

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