UVa 10316 - Airline Hub

题目：给出地球上的n个机场的经度和纬度，想在这里面确定一个HUB使得他到其他机场的最大距离最小。

分析：计算几何、大地坐标系。因为数据不大直接枚举即可，比较时利用圆心角可以提高计算效率，并控制精度。

利用公式可直接解得两点的空间圆心角：acos(cos(lat1)*cos(lat2)*cos(lon1-lon2)+sin(lat1)*sin(lat2))；

因为又是大地坐标，所以再推一遍吧，推导过程如下：

如图，C，D为已知两点则有如下推导：

AB = r*cos(lat1)；DE = r*cos(lat2)；BE = r*sin(lat1) + r*sin(lat2)；

AD*AD = BE*BE + (AB-DE)*(AB-DE) = 2*r*r - 2*r*r*sin(lat1)*sin(lat2) - 2*r*r*cos(lat1)*cos(lat2)；

AC*AC = 2*AB*AB - 2*AB*AB*cos(lon1-lon2) = 2*r*r*cos(lat1)*cos(lat1)*(1-cos(lon1-lon2))；

DF*DF = 2*DE*DE - 2*DE*DE*cos(lon1-lon2) = 2*r*r*cos(lat2)*cos(lat2)*(1-cos(lon1-lon2))；

AC*DF = 2*r*r*cos(lat1)*cos(lat2)*(1-cos(lon1-lon2))；

由托勒密定理有 AC*DF + AD*AD = CD*CD 整理有：

CD = r*sqrt(2-2*(cos(lat1)*cos(lat2)*cos(lon1-lon2)+sin(lat1)*sin(lat2)))；

设圆心角为α则：

cos(0.5*α) = 0.5*CD/r = sqrt(0.5-0.5*(cos(lat1)*cos(lat2)*cos(lon1-lon2)+sin(lat1)*sin(lat2)))；

cos(0.5*α)*cos(0.5*α) = 0.5 - 0.5*(cos(lat1)*cos(lat2)*cos(lon1-lon2)+sin(lat1)*sin(lat2))；

cosα = 1 - 2*cos(0.5*α)*cos(0.5*α) = cos(lat1)*cos(lat2)*cos(lon1-lon2)+sin(lat1)*sin(lat2)；

圆心角为：acos(cos(lat1)*cos(lat2)*cos(lon1-lon2)+sin(lat1)*sin(lat2))；

注意：精度控制。

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cmath>

using namespace std;

double dis[1005][1005];
double lat[1005];
double lon[1005];

double dist( double l1, double d1, double l2, double d2 )
{
	double p = acos(-1.0);
	l1 *= p/180.0; d1 *= p/180.0;
	l2 *= p/180.0; d2 *= p/180.0;
	return acos(cos(l1)*cos(l2)*cos(d1-d2)+sin(l1)*sin(l2));
}

int main()
{
	int n;
	while ( ~scanf("%d",&n) ) {
		for ( int i = 0 ; i < n ; ++ i )
			scanf("%lf%lf",&lat[i],&lon[i]);

		for ( int i = 0 ; i < n ; ++ i )
		for ( int j = i ; j < n ; ++ j )
			dis[i][j] = dis[j][i] = dist( lat[i], lon[i], lat[j], lon[j] );

		double min = 1e20;
		int    spa = 0;
		for ( int i = 0 ; i < n ; ++ i ) {
			double max = 0.0;
			for ( int j = 0 ; j < n ; ++ j )
				if ( max+1e-6 < dis[i][j] )
					max = dis[i][j];
			if ( min+1e-6 > max ) {
				min = max;
				spa = i;
			}
		}

		printf("%.2lf %.2lf\n",lat[spa],lon[spa]);
	}
	return 0;
}