HDU 4766 Network

题意   就是找到一个 位置 使得路由器可以覆盖所有英雄    可以不覆盖主人自己， 找到距离 主人房子最近的位置，距离为多少

没想到  其实是道水题啊！！  分三个情况讨论

第一个情况    如果  把主人的位置放上路由器 可以 覆盖到所有英雄，则答案出来了 0( 一个 半径为 d 的圆，边界上没有点）；

第二个情况    考虑  圆上只有一个点，这个圆只受到该点的约束（ 则圆心在  连线上）；

第三个情况   两个点  或者更多点确定那个圆， 则当两个点或者更多的点都距离为d 时确定那个圆心；因为如果那个点的距离没有到d   证明我还可以更靠近一点房子

 #include<iostream>
 #include<cstring>
 #include<algorithm>
 #include<stdio.h>
 #include<cmath>
 #include<vector>
 #include<functional>
 #define eps 1e-9
 using namespace std;
 const double PI = acos( -1.0 );
 inline int dcmp( double x ){if( abs(x) < eps )return ;else return x<?-:;}
 struct point{
    double x,y;
    point( double x = ,double y =  ):x(x),y(y){}
 }node[];
 typedef point Vector;
 typedef vector<point>ploygon;
 inline point operator+( point a,point b ){ return point(a.x+b.x,a.y+b.y); }
 inline point operator-( point a,point b ){ return point(a.x-b.x,a.y-b.y); }
 inline point operator*( point a,double p){ return point(a.x*p,a.y*p); }
 inline point operator/( point a,double p){ return point(a.x/p,a.y/p); }
 inline bool operator< ( const point &a,const point &b ){return  a.x<b.x||(a.x==b.x&&a.y<b.y);}
 inline bool operator==( const point &a,const point &b ){return (dcmp(a.x-b.x) ==  && dcmp(a.y-b.y) == );}
 inline bool operator!=( const point &a,const point &b ){return a==b?false:true;}

 inline double Dot( point a,point b ){ return a.x*b.x + a.y*b.y; }
 inline double Cross( point a,point b ){ return a.x*b.y - a.y*b.x; }
 inline double Length( point a ){ return sqrt(Dot(a,a)); }
 inline double Angle( point a,point b ){
       double right = Dot(a,b)/Length(a)/Length(b);
       return acos(right);
 }
 inline point Rotate( point a,double rad ){
     return point( a.x*cos(rad)-a.y*sin(rad),a.x*sin(rad+a.y*cos(rad)) );
 }
 point A[]; int N; double dis;
 bool work( point temp )
 {
     for( int i = ; i <= N; i++ )
     if( Length( A[i]-temp ) > dis && abs(Length( A[i] - temp ) - dis) > eps ){
     return false;}
     return true;
 }
 int main( )
 {
     //freopen( "ou.txt","r",stdin);
     //freopen( "in.txt","w",stdout);
     while( scanf("%lf%lf%lf",&A[].x,&A[].y,&dis) != EOF )
     {
         scanf("%d",&N);
         for( int i = ; i <= N; i++ )
         scanf("%lf%lf",&A[i].x,&A[i].y);
            double Max = (<<); bool fell = false;
         for( int i = ; i <= N; i++ )
         if( Length( A[]-A[i] ) > dis )fell = true;
         if( !fell ){ puts("0.00");continue;}
             double Min = (<<);
         for( int i = ; i <= N; i++ ){
             point temp = A[i] + (A[] - A[i])*(dis/Length(A[]-A[i]));
             if( Length( temp-A[] ) < Min ) if( work( temp ) )
             Min = min( Min,Length(temp-A[]) );
         }
         for( int i = ; i <= N; i++ )
         for( int j = i+; j <= N; j++ )
         {
             point temp = (A[i]+A[j])/2.0; double D = Length(A[i]-A[j])*0.5;
              Vector now1 = Rotate( (A[i]-A[j]),PI/2.0 )*sqrt(dis*dis-D*D)/(D*2.0);
              Vector now2 = Rotate( (A[i]-A[j]),-PI/2.0 )*sqrt(dis*dis-D*D)/(D*2.0);
             point temp1 = temp + now1;
             point temp2 = temp + now2;
             if( Length( temp1-A[] ) < Min ) if( work( temp1 ) )
                Min = min( Min,Length(temp1-A[]) );
             if( Length( temp2-A[] ) < Min ) if( work( temp2 ) )
                Min = min( Min,Length(temp2-A[]) );
         }
         if( Min == (<<) )puts("X");
         else printf("%.2lf\n",Min);
     }
     return ;
 }
 /*

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