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给出三角形三个顶点，求出三个互切的圆的半径

虽然大白鼠说可以推出公式，可是这个公式只怕没那么容易推……我左看右看上看下看也推不出。

应该是要做辅助线什么的，那也……

由于很容易就推出了关于三个半径的三元方程组，那么就试试搜索吧，搜其中任意一个半径，只要满足这个方程组就可以了，

那么就二分搜索吧，当然，这个单调性呢？

看图可知，比方说，我们搜最靠近最上面的顶点的圆的半径r1，由于，下面两圆的r2,r3都是由r1推出，因为方程组的约束作用，那么下面两个圆肯定跟最上面的圆相切，当然也肯定跟三角形的边相切，但是这两个圆之间可不一定相切了，如果r1太大，那么这两个圆肯定相离，由它们两推出的底边肯定就小了，如果r1太小，那么这两个圆肯定相交，由它们两推出的底边肯定就大了。好的，有单调性，也就是说，r1跟底边长度是单调递减的关系。

r1的下界就是0，上界却不能太大，因为太大，r2,r3就木有了，就是说方程组是无法通过r1推出满足的r2,r3的，带到方程组里是会出问题的，所以，r1不能太大，那么多大合适呢？列出方程组就很容易知道了。

这是大白鼠的计算几何Basic部分的最后一题，那么这个部分就圆满结束了。通过这个部分，解析几何的能力有所提升。

我的代码：

#include<iostream>
#include<map>
#include<string>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<vector>
#include<algorithm>
using namespace std;
const double eps=1e-7;
struct dot
{double x,y;dot(){}dot(double a,double b){x=a;y=b;}dot operator -(const dot &a){return dot(x-a.x,y-a.y);}double mod(){return sqrt(pow(x,2)+pow(y,2));}double dis(const dot &a) {return sqrt(pow(x-a.x,2)+pow(y-a.y,2));}double mul(const dot &a) {return x*a.x+y*a.y;}
};
int main()
{   bool flag;int i,j;dot a[3],c[3];double l,h,m,b[3],e[3],r[3],A,B,C;while(1){flag=1;for(i=0;i<3;i++){cin>>a[i].x>>a[i].y;if(a[i].x!=0||a[i].y!=0)flag=0;}if(flag)break;for(i=0;i<3;i++){c[0]=a[(i+1)%3]-a[i];c[1]=a[(i+2)%3]-a[i];b[i]=acos(c[0].mul(c[1])/c[0].mod()/c[1].mod())/2;}e[0]=a[0].dis(a[1]);e[1]=a[1].dis(a[2]);e[2]=a[0].dis(a[2]);l=0;h=min(e[0]*tan(b[0]),e[2]*tan(b[0]));while(h-l>eps){m=(l+h)/2;A=1/tan(b[1]);B=2*sqrt(m);C=m/tan(b[0])-e[0];r[0]=(sqrt(B*B-4*A*C)-B)/A/2;A=1/tan(b[2]);B=2*sqrt(m);C=m/tan(b[0])-e[2];r[1]=(sqrt(B*B-4*A*C)-B)/A/2;r[0]*=r[0];r[1]*=r[1];if(r[0]/tan(b[1])+r[1]/tan(b[2])+2*sqrt(r[0]*r[1])-e[1]>0)l=m;elseh=m;}printf("%.6lf %.6lf %.6lf\n",l,r[0],r[1]);}
}

原题：

Time limit: 3.000 seconds

• the circle with the center (25.629089, -10.057956) and the radius 9.942044,
• the circle with the center (53.225883, -0.849435) and the radius 19.150565, and
• the circle with the center (19.701191, 19.203466) and the radius 19.913790.

Your mission is to write a program to calculate the radii of the Malfatti circles of the given triangles.

Input 

The input is a sequence of datasets. A dataset is a line containing six integers x₁, y₁, x₂, y₂, x₃ and y₃ in this order, separated by a space. The coordinates of the vertices of the given triangle are (x₁, y₁), (x₂, y₂) and (x₃, y₃), respectively. You can assume that the vertices form a triangle counterclockwise. You can also assume that the following two conditions hold.

• All of the coordinate values are greater than -1000 and less than 1000.
• None of the Malfatti circles of the triangle has a radius less than 0.1.

The end of the input is indicated by a line containing six zeros separated by a space.

Output 

For each input dataset, three decimal fractions r₁, r₂ and r₃ should be printed in a line in this order separated by a space. The radii of the Malfatti circles nearest to the vertices with the coordinates (x₁, y₁), (x₂, y₂) and (x₃, y₃) should be r₁, r₂ and r₃, respectively.

None of the output values may have an error greater than 0.0001. No extra character should appear in the output.

Sample Input 

20 80 -40 -20 120 -20 
20 -20 120 -20 -40 80 
0 0 1 0 0 1 
0 0 999 1 -999 1 
897 -916 847 -972 890 -925 
999 999 -999 -998 -998 -999 
-999 -999 999 -999 0 731 
-999 -999 999 -464 -464 999 
979 -436 -955 -337 157 -439 
0 0 0 0 0 0

Sample Output 

21.565935 24.409005 27.107493 
9.942044 19.150565 19.913790 
0.148847 0.207107 0.207107 
0.125125 0.499750 0.499750 
0.373458 0.383897 0.100456 
0.706768 0.353509 0.353509 
365.638023 365.638023 365.601038 
378.524085 378.605339 378.605339 
21.895803 22.052921 5.895714