2186: Triangle Centers 三角形的各中心到顶点的距离和

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这是一个程序，用于计算给定三角形的费马点、内心、重心和外心到各个顶点的距离。通过输入三角形的三边长度，程序会输出对应的四个距离。涉及数学和几何知识，包括面积公式和三角函数的运用。

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2186: Triangle Centers

Result	TIME Limit	MEMORY Limit	Run Times	AC Times	JUDGE
	3s	8192K	67	38	Standard

Given the lengths of three sides of a triangle ABC, you are asked to find the following four distances.

(1) FA + FB + FC. F is the point X which minimizes the sum of distances from A, B, and C in XA+XB +XC.

(2) IA + IB + IC. I is the intersection of Angle Bisectors, and it is the interior point for which distances to the sidelines are equal.

(3) GA + GB + GC. G is the intersection of Medians, and it is the centroid of ABC.

(4) OA + OB + OC. O is the intersection of Perpendicular Bisectors, and it is the point for which distances to A, B, C are equal.

Input

The input consists of T test cases. The number of test cases (T ) is given in the first line of the input. Each of the next T lines contains one test case. Each case contains three integer numbers a, b, c where a is the distance between B and C, b is the distance between C and A and c is the distance between A and B. The important condition is 0 < a; b; c <= 1000, and it is guaranteed that the triangle is non-degenerate.

Output

For each set of input you should output four floating-point numbers DF , DI , DG, DO, all of which have exactly three digits after the decimal point. Here
DF = FA + FB + FC
DI = IA + IB + IC
DG = GA + GB + GC
DO = OA + OB + OC

Sample Input

Sample Output

6.766 6.813 6.918 7.500
17.321 17.321 17.321 17.321
9.000 9.316 9.530 14.667

Problem Source: SJTU Programming Contest 2004

#include<stdio.h>
#include<math.h>
#include<algorithm>
using std::swap;
int main()
{
    int T,a,b,c;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d%d",&a,&b,&c);
        if(a>b) swap(a,b);if(b>c) swap(b,c);if(a>c) swap(a,c);
        double s=(a+b+c)*1.0/2;
        double area=sqrt(s*(s-a)*(s-b)*(s-c));
        //费马点
        double l1;
        double d=acos((a*a+b*b-c*c)*1.0/(2*a*b));
        if(d>acos(-1)*2/3) l1=a+b;
        else l1=sqrt(a*a+b*b+c*c+4*sqrt(3.0)*area)*sqrt(2)/2;
        printf("%.3lf ",l1);
        //2 内心，内接圆半径为2*area/(A+B+C),然后用三角公式算下就行。
        double r2=2*area/(a+b+c);
        double t=acos((a*a+b*b-c*c)*1.0/(2*a*b));
        double l2=0;
        l2+=r2/sin(t/2);
        t=acos((a*a+c*c-b*b)*1.0/(2*a*c));
        l2+=r2/sin(t/2);
        t=acos((b*b+c*c-a*a)*1.0/(2*b*c));
        l2+=r2/sin(t/2);
        printf("%.3lf ",l2);
        //重心
        double l3=sqrt((2*a*a+2*b*b-c*c)*1.0/9)+sqrt((2*a*a+2*c*c-b*b)*1.0/9)+sqrt((2*c*c+2*b*b-a*a)*1.0/9);
        printf("%.3lf ",l3);
        //外心
        double r4=a*b*c*1.0/(4*area);
        double l4=3*r4;
        printf("%.3lf/n",l4);
    }
    return 0;
}