Nowadays all circuses in Berland have a round arena with diameter 13 meters, but in the past things were different.
In Ancient Berland arenas in circuses were shaped as a regular (equiangular) polygon, the size and the number of angles could vary from one circus to another. In each corner of the arena there was a special pillar, and the rope strung between the pillars marked the arena edges.
Recently the scientists from Berland have discovered the remains of the ancient circus arena. They found only three pillars, the others were destroyed by the time.
You are given the coordinates of these three pillars. Find out what is the smallest area that the arena could have.
The input file consists of three lines, each of them contains a pair of numbers –– coordinates of the pillar. Any coordinate doesn't exceed 1000 by absolute value, and is given with at most six digits after decimal point.
Output the smallest possible area of the ancient arena. This number should be accurate to at least 6 digits after the decimal point. It's guaranteed that the number of angles in the optimal polygon is not larger than 100.
0.000000 0.000000
1.000000 1.000000
0.000000 1.000000
1.00000000
这道题的题意是： 以一个场地遗迹，呈现多边形，但是不知道具体是几边形，只知道他的三个点，求能包含这三个点的最小多边形的面积：
对于这样的题目： 思路为：
先求出他的外接圆，得到外接圆的半径rr.
(1外接圆的求法：
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（1） 有给定的坐标我们不难求出三条边的边长，rea,reb,rec;
(2) 又海伦公式得到三角形的面积： 周长cc=(rea+reb+rec)/2.0 面积等于： ss=sqrt(cc*(cc-rea)*(cc-reb)*(cc-rec));
(3) rr=rea*reb*rec/(4*ss); //证明就不详细说了
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得到外接园的半径之后：
我们再来求出每一条边对应的圆心角a,b,c;
求出a,b,c圆心角的最大公约数st;
这样我们就可以知道他是边数： 2*pi/st;
所以得到最小单位的三角形的面积为Area=rr*rr*sin(st)/2;
总面积只需再剩上他的边数就可以得到.....
代码如下：
1 #include<iostream> 2 #include<cstdio> 3 #include<cmath> 4 using namespace std; 5 const double PI = 3.1415926535; 6 const double esp=0.01; 7 struct node{ 8 double x,y; 9 //求两点之间的长度 10 double solen(node a){ 11 return sqrt((a.x-x)*(a.x-x)+(a.y-y)*(a.y-y)); 12 } 13 }; 14 double dgcd(double a,double b) //最小公倍数 15 { 16 if(a<esp) return b; 17 if(b<esp) return a; 18 return dgcd(b,fmod(a,b)); 19 } 20 int main() 21 { 22 node a,b,c; 23 double rea,reb,rec,Area; 24 double angle[3]; //角度 25 //freopen("test.in","r",stdin); 26 scanf("%lf%lf%lf%lf%lf%lf",&a.x,&a.y,&b.x,&b.y,&c.x,&c.y); 27 rea=a.solen(b); 28 reb=a.solen(c); 29 rec=b.solen(c); 30 //又海伦公式 31 double cc=(rea+reb+rec)/2.0; 32 Area=sqrt(cc*(cc-rea)*(cc-reb)*(cc-rec)); 33 //求得外接圆半径r 34 double rr=rea*reb*rec/(4*Area); 35 angle[0]=acos(1-rea*rea/(2*rr*rr)); 36 angle[1]=acos(1-reb*reb/(2*rr*rr)); 37 angle[2]=2*PI-angle[0]-angle[1]; 38 //求出角之间的最大公约数 39 double ff=angle[0]; 40 for(int i=1;i<3;i++) 41 ff=dgcd(ff,angle[i]); 42 //求得是多少边行 43 printf("%.6lf ",(rr*rr*PI*sin(ff))/ff); 44 return 0; 45 }