hdu4033(计算几何)

Regular Polygon

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 2905    Accepted Submission(s): 884

Problem Description

In a 2_D plane, there is a point strictly in a regular polygon with N sides. If you are given the distances between it and N vertexes of the regular polygon, can you calculate the length of reguler polygon's side? The distance is defined as dist(A, B) = sqrt( (Ax-Bx)*(Ax-Bx) + (Ay-By)*(Ay-By) ). And the distances are given counterclockwise.

 

Input

First a integer T (T≤ 50), indicates the number of test cases. Every test case begins with a integer N (3 ≤ N ≤ 100), which is the number of regular polygon's sides. In the second line are N float numbers, indicate the distance between the point and N vertexes of the regular polygon. All the distances are between (0, 10000), not inclusive.

 

Output

For the ith case, output one line “Case k: ” at first. Then for every test case, if there is such a regular polygon exist, output the side's length rounded to three digits after the decimal point, otherwise output “impossible”.

 

Sample Input

233.0 4.0 5.031.0 2.0 3.0

 

Sample Output

Case 1: 6.766Case 2: impossible

 

Source

The 36th ACM/ICPC Asia Regional Chengdu Site —— Online Contest

 

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      本题给定正多边形内一点到各个顶点的距离，要求是否存在这样的多边形。

      本题的突破口是多边形为正多边形，几个边相等，对应的内角相等。我们可以二分枚举边长，只要同时满足下面的两个条件即可：

 

 1.构成三角形

 2.角度等于360.

      本题在卡精度，不知为什么一直WA。

#include<iostream>#include<cmath>#include<cstdio>using namespace std;const double PI2=2*acos(-1.0);const double eps=1e-8;const int MAXN=100+10;double Edg[MAXN];int n;double Angle(double a,double b,double c){return acos((a*a+b*b-c*c)/(2.0*a*b));}int IsPology(double len){double sumangle=0.0;for(int i=1;i<n+1;i++){sumangle+=Angle(Edg[i],Edg[i-1],len);}if(fabs(sumangle-PI2)<=eps)return 0;else if(sumangle>PI2)return -1;return 1;}void Solve(){double high,low,mid;Edg[n]=Edg[0];high=Edg[0]+Edg[1];low=fabs(Edg[0]-Edg[1]);for(int i=2;i<n+1;i++){double tmp=Edg[i]+Edg[i-1];if(high<tmp)high=tmp;tmp=fabs(Edg[i]-Edg[i-1]);if(tmp>low)low=tmp;}bool flag=false;high-=eps;low+=eps;while(high>low+eps){mid=(high+low)/2.0;int t=IsPology(mid);if(t==0){flag=true;break;}else if(t==-1)high=mid;else low=mid;}if(flag)printf("%0.3lf\n",mid);else printf("impossible\n");}int main(){int cas;int tag=1;scanf("%d",&cas);while(cas--){scanf("%d",&n);for(int i=0;i<n;i++)scanf("%lf",&Edg[i]);printf("Case %d: ",tag++);Solve();}system("pause");return 0;}