HDU-4720 Naive and Silly Muggles 圆的外心

　　题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4720

　　先两两点之间枚举，如果不能找的最小的圆，那么求外心即可。。

//STATUS:C++_AC_0MS_292KB
#include <functional>
#include <algorithm>
#include <iostream>
//#include <ext/rope>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <cstring>
#include <cassert>
#include <cstdio>
#include <string>
#include <vector>
#include <bitset>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <list>
#include <set>
#include <map>
using namespace std;
//#pragma comment(linker,"/STACK:102400000,102400000")
//using namespace __gnu_cxx;
//define
#define pii pair<int,int>
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define PI acos(-1.0)
//typedef
typedef __int64 LL;
typedef unsigned __int64 ULL;
//const
const int N=25;
const int INF=0x3f3f3f3f;
const int MOD=1000000007,STA=8000010;
const LL LNF=1LL<<60;
const double EPS=1e-8;
const double OO=1e60;
const int dx[4]={-1,0,1,0};
const int dy[4]={0,1,0,-1};
const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
//Daily Use ...
inline int sign(double x){return (x>EPS)-(x<-EPS);}
template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
template<class T> inline T Min(T a,T b){return a<b?a:b;}
template<class T> inline T Max(T a,T b){return a>b?a:b;}
template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
//End

int T;

struct Point{
    double x,y;
}p[5];

double dist(Point a,Point b)
{
    return sqrt( (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y) );
}

double sqr(double x){ return x * x; }

void Ci(Point p0 , Point p1 , Point p2 , Point &cp)
{
    double a1=p1.x-p0.x,b1 = p1.y - p0.y,c1 = (sqr(a1) + sqr(b1)) / 2;
    double a2=p2.x-p0.x,b2 = p2.y - p0.y,c2 = (sqr(a2) + sqr(b2)) / 2;
    double d = a1 * b2 - a2 * b1;
    cp.x = p0.x + (c1 * b2 - c2 * b1) / d;
    cp.y = p0.y + (a1 * c2 - a2 * c1) / d;
}

int judge1(int& ok)
{
    int i,j,k;
    double d,r,lowr;
    Point c,t;
    lowr=OO;
    for(i=0;i<3;i++){
        for(j=0;j<3;j++){
            if(i==j)continue;
            r=dist(p[i],p[j])/2;
            t.x=(p[i].x+p[j].x)/2;
            t.y=(p[i].y+p[j].y)/2;
            for(k=0;k==i || k==j;k++);
            if(dist(t,p[k])<=r){
                lowr=r;
                c.x=t.x,c.y=t.y;
            }
        }
    }
    if(lowr==OO)return 0;
    if(dist(p[3],c)<=lowr)ok=1;
    return 1;
}

int main(){
 //   freopen("in.txt","r",stdin);
    int i,j,ca=1,ok;
    double r,d;
    Point c;
    scanf("%d",&T);
    while(T--)
    {
        for(i=0;i<4;i++){
            scanf("%lf%lf",&p[i].x,&p[i].y);
        }
        ok=0;
        if(judge1(ok));
        else {
            Ci(p[0],p[1],p[2],c);
            r=(c.x-p[0].x)*(c.x-p[0].x)+(c.y-p[0].y)*(c.y-p[0].y);
            d=(c.x-p[3].x)*(c.x-p[3].x)+(c.y-p[3].y)*(c.y-p[3].y);
            if(d<=r)ok=1;
        }

        printf("Case #%d: %s
",ca++,ok?"Danger":"Safe");
    }
    return 0;
}