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#include #include #include 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;