Maple trees（最小覆蓋圓）

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Maple trees

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

Total Submission(s): 222 Accepted Submission(s): 79

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 ?

InputThe 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.

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

Sample Input 21 0-1 00

Sample Output 1.50

Authorzjt

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/*題意：給你散落的點，讓你求出最小的圓，將這些點圍起來，點可以在圓上，輸出圓的最小半徑。初步思路：求出凸包，然後在求出這個凸包的外接圓，兩條邊的垂直平分線的交點就是圓心#錯誤：有點天真，三角形一定有外接圓，但是多邊形不一定有外接圓#改進：求出凸包，然後求凸包的最小覆蓋圓，這個名詞也是看到部落格才知道了（嗚嗚嗚，又少了       一次好好動腦的機會）然後任意的三個點組成的三角形的外接圓的最大半徑就是就是凸包       “外接圓”的半徑了，三角形的外接圓半徑為abc/4s，這個公式能簡單的證明。遍曆出所有的半徑，中間最長的半徑       就是要求的半徑。#改進錯誤點：上面的情況只適用於銳角三角形，鈍角，直角三角形的“外接圓”的最小半徑，不是外接圓的半徑，而是最長邊的一半！       */#include<bits/stdc++.h>using namespace std;/****************************凸包模板*******************************/const double eps = 1e-8;int sgn(double x){    if(fabs(x) < eps)return 0;    if(x < 0)return -1;    else return 1;}struct Point{    double x,y;    Point(){}    Point(double _x,double _y)    {        x = _x;y = _y;    }    Point operator -(const Point &b)const    {        return Point(x - b.x,y - b.y);    }    //叉積    double operator ^(const Point &b)const    {        return x*b.y - y*b.x;    }    //點積    double operator *(const Point &b)const    {        return x*b.x + y*b.y;    }    void input(){        scanf("%lf%lf",&x,&y);    }};struct Line {     Point s,e;     Line(){}     Line(Point _s,Point _e) {                 s = _s; e = _e;     }}; //*兩點間距離double dist(Point a,Point b){    return sqrt((a-b)*(a-b));}   /** 求凸包，Graham演算法* 點的編號0~n-1* 返回凸包結果Stack[0~top-1]為凸包的編號*/const int MAXN = 105;Point List[MAXN];int Stack[MAXN];//用來存放凸包的點int top;//表示凸包中點的個數//相對於List[0]的極角排序bool _cmp(Point p1,Point p2){    double tmp = (p1-List[0])^(p2-List[0]);    if(sgn(tmp) > 0)        return true;    else if(sgn(tmp) == 0 && sgn(dist(p1,List[0]) - dist(p2,List[0])) <= 0)        return true;    else         return false;}void Graham(int n){    Point p0;    int k = 0;    p0 = List[0];    //找最下邊的一個點    for(int i = 1;i < n;i++)    {        if( (p0.y > List[i].y) || (p0.y == List[i].y && p0.x > List[i].x) )        {            p0 = List[i];            k = i;        }    }    swap(List[k],List[0]);    sort(List+1,List+n,_cmp);    if(n == 1)    {        top = 1;        Stack[0] = 0;        return;    }    if(n == 2)    {        top = 2;        Stack[0] = 0;        Stack[1] = 1;        return ;    }    Stack[0] = 0;    Stack[1] = 1;    top = 2;    for(int i = 2;i < n;i++)    {        while(top > 1 && sgn((List[Stack[top-1]]-List[Stack[top-2]])^(List[i]-List[Stack[top-2]])) <= 0)            top--;        Stack[top++] = i;    }}/****************************凸包模板*******************************/int n;int main(){    // freopen("in.txt","r",stdin);    while(scanf("%d",&n)!=EOF&&n){        for(int i=0;i<n;i++){            List[i].input();        }//輸入所有點座標        if(n==1){            printf("0.50\n");            continue;        }        if(n==2){            printf("%.2lf\n",dist(List[0],List[1])/2+0.5);            continue;        }        Graham(n);//求出凸包        double Maxr=-1.0;        // cout<<top<<endl;        //將Static[0]作為所有小三角形的公用頂點        for(int i=0;i<top;i++){//枚舉三角形的點            for(int j=i+1;j<top;j++){                for(int k=j+1;k<top;k++){                    /*                    三條邊的長度                    */                    double a=dist(List[Stack[i]],List[Stack[j]]);                    double b=dist(List[Stack[i]],List[Stack[k]]);                    double c=dist(List[Stack[k]],List[Stack[j]]);                    if(a*a+b*b<c*c||a*a+c*c<b*b||b*b+c*c<a*a){//判斷是不是銳角三角形                        Maxr=max(Maxr,max(max(a,b),c)/2);                    }else{                        /*                        三角形的面積                        */                        Point x1=List[Stack[j]]-List[Stack[i]];                        Point x2=List[Stack[k]]-List[Stack[i]];                        double s=fabs(x1^x2)/2;                        Maxr=max(Maxr,(a*b*c)/(4*s));                    }                }            }        }        printf("%.2lf\n",Maxr+0.5);    }    return 0;}

 

Maple trees（最小覆蓋圓）