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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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