Network（计算几何：推规律+暴力+贪心）

Link：http://acm.hdu.edu.cn/showproblem.php?pid=4766

Network

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 713    Accepted Submission(s): 171

Problem Description

With the help of the Ares L, the Emperor of Law and some stars of the future, General M succeeded in setting foot on the mountain, and had set up a kingdom, known as ACMSYSU. In order to reward these founding heroes, General M decided to share her network.
General M has manufactured a router. One can use this network if the Euclidean distance between his/her house and the router is not greater than d. Because of General M's excellent technology, the distance between her house and the router can be infinite.
There are n heroes in the kingdom besides General M. General M is not an eccentric person, so she hopes everyone can use the network. Because the farther the router is away from her house, the worse the signal, General M wants the router being as close to her house as possible. Please help General M place the router so that the distance between her house and the router is minimized, and every one can use the router. You just need to tell General M the minimum distance.

 

Input

The input will consist of multiple test cases. 
For each case, the first line contains 3 integers x, y and d, which represent that the General M's house locates at the point (x, y) and one can use the network if his/her distance to the router is not greater than d.
The second line is a positive integer n, the number of heroes of the kingdom.
Then n lines follow. Each line has 2 integers x_i and y_i, indicating that the i-th hero locates at point (x_i, y_i)。

All integers in the input satisfy that their absolute values are less than or equal to 10^5, and 1<=n<=10^3.

 

Output

For each case, output a real number which indicates the minimum distance between the General M's house and the router, to two decimal places.
If it is impossible to share a network which can cover everyone, output an 'X' (without the single quote).

 

Sample Input

0 0 525 35 -30 0 125 35 -3

 

Sample Output

1.00X

 

Source

2013 ACM/ICPC Asia Regional Changchun Online

 

以下分析来自：http://www.cnblogs.com/wulangzhou/archive/2013/09/30/3346578.html

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

AC code：

#include<iostream>#include<cstring>#include<algorithm>#include<stdio.h>#include<cmath>#include<vector>#define eps 1e-9using namespace std;const double PI=acos( -1.0 );inline int dcmp( double x ){if( abs(x) < eps )return 0;else return x<0?-1:1;}struct point{   double x,y;   point( double x = 0,double y = 0 ):x(x),y(y){}}node[112345];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) == 0 && dcmp(a.y-b.y) == 0);}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[1123]; int N; double dis;bool work( point temp ){    for( int i = 1; i <= N; i++ )    if( Length( A[i]-temp ) > dis && abs(Length( A[i] - temp ) - dis) > eps ){    return false;}    return true;}int main(){while( scanf("%lf%lf%lf",&A[0].x,&A[0].y,&dis) !=EOF){scanf("%d",&N);for(int i=1;i<=N;i++){scanf("%lf%lf",&A[i].x,&A[i].y);}if(work(A[0])){ puts("0.00");}else{double Min=(1<<30);for(int i=1;i<=N;i++){point temp=A[i]+(A[0]-A[i])*(dis/Length(A[0]-A[i]));//圆心 if(Min>Length(temp-A[0])){if(work(temp)){Min=Length(temp-A[0]);}}}for(int i=1;i<=N;i++){for(int j=i+1;j<=N;j++){point temp = (A[i]+A[j])/2.0;//线段中点 double D = Length(A[i]-A[j])*0.5;//线段一半的长度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);//过两点满足圆心与这两点距离为dis的圆心有两个 point temp1=temp+now1;point temp2=temp+now2;    if( Length( temp1-A[0] ) < Min ){if( work( temp1 ) ){Min=Length( temp1-A[0] );}}if(Length( temp2-A[0] ) < Min){if( work( temp2 ) ){Min=Length( temp2-A[0] );}} }}if( Min==(1<<30) ) puts("X");else printf("%.2lf\n",Min);}}return 0;}