uva 11524 - InCircle (二分法)

题意：三角形ABC的内切圆把它的三边分别划分成 m1：n1，m2：n2 和 m3：n3 的比例。另外已知内切圆的半径 r ，求三角形ABC 的面积。

#include<iostream>

#include<iomanip>

#include<algorithm>

#include<cmath>

#define sqr(a) (a)*(a)

#define eps 1e-12

#define min(a,b) a<b?a:b

#define max(a,b) a>b?a:b

#define pi asin(1.0)

using namespace std;

int sig(double a)

{

    return (a>eps)-(a<-eps);

}

int main()

{

    int t;

    double r,m1,n1,m2,n2,m3,n3,k1,k2,k3;

    double left,right,mid,thy,th1,th2,th3;

    cin>>t;

    while(t--)

    {

        cin>>r>>m1>>n1>>m2>>n2>>m3>>n3;

        k1=sqr(n1);

        k2=sqr(n2/m2)*k1;

        k3=sqr(m1);

        left=min(sqrt(3/k1)*r,sqrt(3/k2)*r);

        left=min(left,sqrt(3/k3)*r);

        right=max(sqrt(3/k1)*r,sqrt(3/k2)*r);

        right=max(right,sqrt(3/k3)*r);

        mid=(left+right)/2;

        while(sig(right-left)>0)

        {

            th1=r/sqrt(k1*sqr(mid)+sqr(r));

            th2=r/sqrt(k2*sqr(mid)+sqr(r));

            th3=r/sqrt(k3*sqr(mid)+sqr(r));

            thy=asin(th1)+asin(th2)+asin(th3);

            int f=sig(thy-pi);

            if(f==0) break;

            else if(f<0) right=mid;

            else left=mid;

            mid=(left+right)/2;

        }

        thy=2*asin(r/sqrt(k1*sqr(mid)+sqr(r)));

        double area=(n1+m1)*mid/2*(n2+m2)*n1/m2*mid*sin(thy);

        cout<<fixed<<setprecision(4)<<area<<endl;

    }

    return 0;

}