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当是钝角三角形时,就是最长边为直径的圆最小.

否则,求三角形的外接圆

Naive and Silly Muggles

3
0 0
2 0
1 2
1 -0.5

0 0
2 0
1 2
1 -0.6

0 0
3 0
1 1
1 -1.5

Case #1: Danger
Case #2: Safe
Case #3: Safe

/* ***********************************************
Author        :CKboss
Created Time  :2015年01月07日 星期三 23时26分30秒
File Name     :HDOJ4720.cpp
************************************************ */

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>

using namespace std;

double eps = 1e-10;

int dcmp(double x) { if(fabs(x)<eps) return 0; return (x<0)?-1:1;}

struct Point
{
    double x,y;
    Point(double _x,double _y):x(_x),y(_y){}
    Point(){}
};

Point operator-(Point A,Point B) {return Point(A.x-B.x,A.y-B.y);}

struct Circle
{
    Point c;
    double r;
    Circle(Point _c,double _r):c(_c),r(_r){}
};

double Dot(Point A,Point B) { return A.x*B.x+A.y*B.y; }
double Length(Point A) { return sqrt(Dot(A,A)); }

Circle CircumscribedClircle(Point p1,Point p2,Point p3)
{
    double Bx=p2.x-p1.x,By=p2.y-p1.y;
    double Cx=p3.x-p1.x,Cy=p3.y-p1.y;
    double D=2*(Bx*Cy-By*Cx);
    double cx=(Cy*(Bx*Bx+By*By)-By*(Cx*Cx+Cy*Cy))/D+p1.x;
    double cy=(Bx*(Cx*Cx+Cy*Cy)-Cx*(Bx*Bx+By*By))/D+p1.y;
    Point p = Point(cx,cy);
    return Circle(p,Length(p1-p));
}

Point p1,p2,p3,pp;

bool check(Circle c , Point p)
{
    double l1 = Length(c.c-p);
    double l2 = c.r;

    if(dcmp(l1-l2)<=0) return false;
    return true;
}

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);

    int T_T,cas=1;
    scanf("%d",&T_T);
    while(T_T--)
    {
        double x,y;
        scanf("%lf%lf",&x,&y);
        p1=Point(x,y);
        scanf("%lf%lf",&x,&y);
        p2=Point(x,y);
        scanf("%lf%lf",&x,&y);
        p3=Point(x,y);
        scanf("%lf%lf",&x,&y);
        pp=Point(x,y);

        // two point on the magic circle

        printf("Case #%d: ",cas++);
        /// p1 ... p2
        double rrr = Length(p1-p2)/2;
        Point ppp = Point((p1.x+p2.x)/2,(p1.y+p2.y)/2);
        Circle ccc = Circle(ppp,rrr);
        if(check(ccc,p3)==false)
        {
            if(check(ccc,pp)==true) puts("Safe");
            else puts("Danger");
            continue;
        }
        /// p2...p3
        rrr = Length(p2-p3)/2;
        ppp = Point((p2.x+p3.x)/2.,(p2.y+p3.y)/2.);
        ccc = Circle(ppp,rrr);
        if(check(ccc,p1)==false)
        {
            if(check(ccc,pp)==true) puts("Safe");
            else puts("Danger");
            continue;
        }
        /// p1...p3
        rrr = Length(p1-p3)/2;
        ppp = Point((p1.x+p3.x)/2.,(p1.y+p3.y)/2.);
        ccc = Circle(ppp,rrr);
        if(check(ccc,p2)==false)
        {
            if(check(ccc,pp)==true) puts("Safe");
            else puts("Danger");
            continue;
        }


        /// three point on circle
        Circle cc = CircumscribedClircle(p1,p2,p3);
        double len = Length(cc.c-pp);
        if(dcmp(len-cc.r)<=0) puts("Danger");
        else puts("Safe");
    }
    
    return 0;
}

HDOJ 4720 Naive and Silly Muggles 三角形外接圆

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原文地址：http://blog.csdn.net/ck_boss/article/details/42521299