HDU 3007 Buried memory 最小圆覆盖

来源：互联网发布：单片机如何检测电压编辑：程序博客网时间：2024/04/30 03:25

题目大意：没看。反正就是求最小圆覆盖。

思路：一个神奇的算法——随机增量法。可以证明，这个算法可以在O(n)的时间复杂度内求出最小圆覆盖。虽然好像能卡掉的样子，但是加上一句random_shuffle就卡不掉了。

具体的过程是这样的：

在全局记录一个圆，表示目前的最小圆覆盖。从头开始扫描。遇到第一个不在当前最小圆覆盖内的点的时候：

将这个点与当前最小圆覆盖的圆心为直径做一个圆，作为当前的最小圆覆盖。从头开始扫描。遇到第一个不在当前最小圆覆盖的点的时候：

将刚才的两个点和当前点做三角形，将这个三角形的外接圆作为当前的最小圆覆盖。

具体还是看代码吧，关于证明见：http://blog.csdn.net/lthyxy/article/details/6661250

CODE：

#define _CRT_SECURE_NO_WARNINGS#include <cmath>#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#define MAX 510using namespace std;struct Point{double x,y;Point(double _,double __):x(_),y(__) {}Point() {}Point operator +(const Point &a)const {return Point(x + a.x,y + a.y);}Point operator -(const Point &a)const {return Point(x - a.x,y - a.y);}Point operator *(double a)const {return Point(x * a,y * a);}void Read() {scanf("%lf%lf",&x,&y);}}point[MAX];inline double Calc(const Point &p1,const Point &p2){return sqrt((p1.x - p2.x) * (p1.x - p2.x) + (p1.y - p2.y) * (p1.y - p2.y));}inline Point Mid(const Point &p1,const Point &p2){return Point((p1.x + p2.x) / 2,(p1.y + p2.y) / 2);}inline double Cross(const Point &p1,const Point &p2){return p1.x * p2.y - p1.y * p2.x;}inline Point Change(const Point &v){return Point(-v.y,v.x);}struct Circle{Point o;double r;Circle(const Point &_,double __):o(_),r(__) {}Circle() {}bool InCircle(const Point &p) {return Calc(p,o) <= r;}}now;struct Line{Point p,v;Line(const Point &_,const Point &__):p(_),v(__) {}Line() {}};inline Point GetIntersection(const Line &l1,const Line &l2){Point u = l1.p - l2.p;double t = Cross(l2.v,u) / Cross(l1.v,l2.v);return l1.p + l1.v * t;}int points;int main(){while(scanf("%d",&points),points) {for(int i = 1; i <= points; ++i)point[i].Read();random_shuffle(point + 1,point + points + 1);now = Circle(point[1],.0);for(int i = 2; i <= points; ++i)if(!now.InCircle(point[i])) {now = Circle(point[i],.0);for(int j = 1; j < i; ++j)if(!now.InCircle(point[j])) {now = Circle(Mid(point[i],point[j]),Calc(point[i],point[j]) / 2);for(int k = 1; k < j; ++k)if(!now.InCircle(point[k])) {Line l1(Mid(point[i],point[j]),Change(point[j] - point[i]));Line l2(Mid(point[j],point[k]),Change(point[k] - point[j]));Point intersection = GetIntersection(l1,l2);now = Circle(intersection,Calc(point[i],intersection));}}}printf("%.2lf %.2lf %.2lf\n",now.o.x,now.o.y,now.r);}return 0;}

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