ZOJ 1598 Spherical Triangle

题意：给出经纬度，求球面三角形的三角和。

看了一下午的维基百科才知道他到底说了什么，英语是真烂啊。。。晚上A码的时候也特别慢，看来自己修炼之路还很长，最后还是在师哥的协助下A掉

公式： $\cos c= \cos a \cos b + \sin a \sin b \cos C. \!$ (其中a,b,c都是弧度，C是球面三角形的一个角)

View Code

#include <stdio.h>
#include <math.h>
const double PI=acos(-1);
struct Node
{
    double x,y,z;
}node[4];

double dis(Node A,Node B)
{
    return (A.x-B.x)*(A.x-B.x)+(A.y-B.y)*(A.y-B.y)+(A.z-B.z)*(A.z-B.z);
}
int main(int argc, char *argv[])
{
    double a_2,b_2,c_2,x,y;
    int T,i;
    double A1,B1,C1,A2,B2,C2;
    scanf("%d",&T);
    while(T--)
    {
        for (i=0;i<3;i++)
        { 
           scanf("%lf%lf",&x,&y);
           node[i].x=cos(x*PI/180.0)*cos(y*PI/180.0);
           node[i].y=cos(y*PI/180.0)*sin(x*PI/180.0);
           node[i].z=sin(y*PI/180.0);
        }
        a_2=dis(node[0],node[1]); 
        b_2=dis(node[1],node[2]); 
        c_2=dis(node[0],node[2]);
        A1=acos(1-a_2/2.0);  
        B1=acos(1-b_2/2.0);  
        C1=acos(1-c_2/2.0);  
        A2=acos((cos(A1)-cos(B1)*cos(C1))/(sin(C1)*sin(B1))); 
        B2=acos((cos(B1)-cos(A1)*cos(C1))/(sin(A1)*sin(C1))); 
        C2=acos((cos(C1)-cos(A1)*cos(B1))/(sin(A1)*sin(B1))); 
        printf("%.2lf\n",(A2+B2+C2)*180.0/PI);
    }
    return 0;
}

附：

$\frac{\sin a}{\sin A}=\frac{\sin b}{\sin B}=\frac{\sin c}{\sin C}.$

$\tan\left(\frac{c}{2}\right) = \sqrt{\frac {-\cos S \cos(S-C)}{\cos (S-A) \cos (S-B)}} \,,$

where $S = (A +B + C)/2$ .