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=;
 const int INF=0x3f3f3f3f;
 const int MOD=,STA=;
 const LL LNF=1LL<<;
 const double EPS=1e-;
 const double OO=1e60;
 const int dx[]={-,,,};
 const int dy[]={,,,-};
 const int day[]={,,,,,,,,,,,,};
 //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[];

 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)) / ;
     double a2=p2.x-p0.x,b2 = p2.y - p0.y,c2 = (sqr(a2) + sqr(b2)) / ;
     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=;i<;i++){
         for(j=;j<;j++){
             if(i==j)continue;
             r=dist(p[i],p[j])/;
             t.x=(p[i].x+p[j].x)/;
             t.y=(p[i].y+p[j].y)/;
             for(k=;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 ;
     if(dist(p[],c)<=lowr)ok=;
     return ;
 }

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

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