簡單的幾何題，先把經(jīng)緯度換算成球面坐標(biāo)，再把球面坐標(biāo)換算成直角坐標(biāo)，然后求夾角，乘半徑得到球面距離
/**//*************************************************************************
Author: WHU_GCC
Created Time: 2007-10-2 10:18:27
File Name: pku3407.cpp
Description:
************************************************************************/
#include <iostream>
#include <cmath>
using namespace std;
#define out(x) (cout<<#x<<": "<<x<<endl)
const int maxint=0x7FFFFFFF;
typedef long long int64;
const int64 maxint64 = 0x7FFFFFFFFFFFFFFFLL;
template<class T>void show(T a, int n)
{for(int i=0; i<n; ++i) cout<<a[i]<<' '; cout<<endl;}
template<class T>void show(T a, int r, int l){for(int i=0; i<r; ++i)show(a[i],l);cout<<endl;}
const double pi = acos(-1.0);
int main()
{
double theta, phi, theta_m, phi_m;
double theta1, phi1, theta2, phi2;
char s1[10], s2[10];
while (scanf("%lf%lf%s%lf%lf%s", &theta, &theta_m, s1, &phi, &phi_m, s2) != EOF)
{
theta += theta_m / 60;
phi += phi_m / 60;
if (strcmp(s1, "N") == 0)
theta1 = 90 - theta;
else
theta1 = 90 + theta;
if (strcmp(s2, "E") == 0)
phi1 = phi;
else
phi1 = 360 - phi;
scanf("%lf%lf%s%lf%lf%s", &theta, &theta_m, s1, &phi, &phi_m, s2);
theta += theta_m / 60;
phi += phi_m / 60;
if (strcmp(s1, "N") == 0)
theta2 = 90 - theta;
else
theta2 = 90 + theta;
if (strcmp(s2, "E") == 0)
phi2 = phi;
else
phi2 = 360 - phi;
theta1 = theta1 * pi / 180;
theta2 = theta2 * pi / 180;
phi1 = phi1 * pi / 180;
phi2 = phi2 * pi / 180;
double p = 6370.0;
double x1, y1, z1, x2, y2, z2;
x1 = p * sin(theta1) * cos(phi1);
y1 = p * sin(theta1) * sin(phi1);
z1 = p * cos(theta1);
x2 = p * sin(theta2) * cos(phi2);
y2 = p * sin(theta2) * sin(phi2);
z2 = p * cos(theta2);
double alpha = acos((x1 * x2 + y1 * y2 + z1 * z2) / ((sqrt(x1 * x1 + y1 * y1 + z1 * z1)) * (sqrt(x2 * x2 + y2 * y2 + z2 * z2))));
printf("%.3lf\n", alpha * p);
}
return 0;
}