ZOJ 3598 Spherical Triangle (三角关系)

ZOJ Problem Set - 3598

Spherical Triangle

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Time Limit: 2 Seconds       Memory Limit: 65536 KB

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As everybody knows, the sum of the interior angles of a triangle on a plane is always 180 degree. But this is not true when the triangle is on spherical surface. Given a triangle on a spherical surface, you are asked to calculate the sum of the interior angles of the triangle.

Formally, you are given the 3 vertex of the triangle. They are connected by the arcs of the great circles, i.e. circles whose centers coincide with the center of the sphere. It is guaranteed that the triangle is not degenerate, i.e. the 3 vertices will not lie on one great circle and no two vertices collide. The interior of the triangle is defined as the smaller part that the triangle is divide into.

Input

There are multiple test cases. The first line of input contains an integer T (0 < T ≤ 2012) indicating the number of test cases. Then T test cases follow.

Each test case contains 3 lines, indicating the position of the 3 vertices. Each line contains 2 real number, each of which contains at most 2 digits after the decimal point, indicating the longitude and the latitude of the vertex. The longitude and the latitude are measured in degree. The longitude will be in (-180, 180] while the latitude will be in [-90, 90].

Output

For each test case, output the sum of the interior angles of the triangle measured in degree, accurate to 0.01.

Sample Input

1

0 0

90 0

0 90

Sample Output

270.00

References

• http://en.wikipedia.org/wiki/Spherical_trigonometry

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Author:  GUAN, Yao
Contest:  The 12th Zhejiang University Programming Contest

 

题意：给出三个点的经度和纬度，求球面三角形的内角和。

分析：根据题目给出的链接的提示结合三角函数计算推出公式就行。

#include<cstdio>

#include<iostream>

#include<cstring>

#include<cmath>

#include<stdlib.h>

#include<vector>

#include<queue>

#include<stack>

#include<algorithm>

using namespace std;

const double PI=acos(-1.0);

struct node

{

    double x;

    double y;

    double z;

}A,B,C;



void change(double j,double w,node &P)

{

    P.x=cos(w*PI/180)*cos(j*PI/180);

    P.y=cos(w*PI/180)*sin(j*PI/180);

    P.z=sin(w*PI/180);

}



double calc(node PA,node PB)

{

    return (PA.x*PB.x)+(PA.y*PB.y)+(PA.z*PB.z);

}



int main()

{

    int kase;

    scanf("%d",&kase);

    while(kase--)

    {

        double ax,ay,bx,by,cx,cy;

        scanf("%lf %lf %lf %lf %lf %lf",&ax,&ay,&bx,&by,&cx,&cy);

        change(ax,ay,A);

        change(bx,by,B);

        change(cx,cy,C);

        //printf("%lf %lf %lf\n",A.x,A.y,A.z);

        double a,b,c;

        a=acos(calc(B,C));

        b=acos(calc(A,C));

        c=acos(calc(A,B));

        //printf("a=%lf b=%lf c=%lf\n",a,b,c);



        double biga,bigb,bigc;

        biga=acos( ( cos(a)-cos(b)*cos(c) )/ ( sin(b)*sin(c) ) );

        bigb=acos( ( cos(b)-cos(a)*cos(c) )/ ( sin(a)*sin(c) ) );

        bigc=acos( ( cos(c)-cos(a)*cos(b) )/ ( sin(a)*sin(b) ) );



        double ans=biga+bigb+bigc;

        //printf("%.2lf\n",ans);

        ans=ans*(180/PI);

        printf("%.2lf\n",ans);

    }

    return 0;

}

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