The Best Polygon

An n-gon is a polygon with n sides. For example, a triangle is a 3-gon. Now you are asked to find the best n-gon in a given convex N-gon. The vertices of the n-gon are selected from vertices of the N-gon. The n-gon you are supposed to find must have the largest area among all possible n-gons, which means that it approximates the N-gon the best. The figure below shows the best 6-gon (the shaded part) in a 10-gon.

Input Specification:

Each input file contains one test case. For each case, the first line gives two positive integers N ( 6 ) which is the total number of vertices of the convex N-gon, and n ( 6 ) which is the total number of vertices of the approximating convex n-gon. Then N lines follow, the i-th line gives the 2-D coordinates (x, y) of the i-th vertex ( , ). The x and y coordinates in a line are real numbers with their absolute values no more than 1000, and they are separated by a space.

Output Specification:

Print in a line all the vertex indices of the best n-gon in descending order. All the numbers are separated by a space and there must be no extra space at the beginning or the end of the line. It is guaranteed that the solution is unique.

Sample Input:

10 6
133.0 1.0
544.0 71.0
558.0 206.0
536.0 338.0
463.0 436.0
330.0 503.0
188.0 499.0
305.0 2.0
55.0 410.0
2.0 140.0

 

Sample Output:

9 8 5 3 1 0

  1 #include
  2 using namespace std;
  3 double x,y;
  4 int n,m;
  5 const double eps=1e-8;
  6 inline int sgn(double d)
  7 {return d<-eps?-1:d>eps;}
  8 struct point
  9 {
 10     double x;
 11     double y;
 12     int id;
 13 } po;
 14 struct cmp1 
 15 {
 16     bool operator() (const point& p1,const point& p2) const
 17     {
 18         return atan2((p1.y-y),(p1.x-x))x));
 19     }
 20 };
 21 vector v;
 22 vector<double> va;
 23 inline double area(int n1,int n2)
 24 {
 25     if(!sgn(va[n1*n+n2]))
 26     {
 27         va[n1*n+n2]=v[n1].x*v[n2].y-v[n1].y*v[n2].x;
 28         va[n2*n+n1]=-va[n1*n+n2];
 29     }
 30     return va[n1*n+n2];
 31 }
 32 int main()
 33 {
 34 //    freopen("data1.txt","r",stdin);
 35     scanf("%d %d",&n,&m);
 36     va.resize(n*n,0.0);
 37     x=0.0;
 38     y=0.0;
 39     for(int i=0;i)
 40     {
 41         scanf("%lf %lf",&po.x,&po.y);
 42         x+=po.x;
 43         y+=po.y;
 44         po.id=i;
 45         v.emplace_back(po);
 46     }
 47     x=x/n;
 48     y=y/n;
 49     sort(v.begin(),v.end(),cmp1());
 50     vectordouble,int> > vdp((n-2)*(n-2)*(m-2),make_pair(0.0,0));
 51     for(int i=0;i2;i++)
 52     {
 53         for(int j=i+2;j)
 54         {
 55             double maxs=0.0;
 56             int k=i+1;
 57             for(;k<j;)
 58             {
 59                 double s=area(i,k)+area(k,j)+area(j,i);
 60                 if(sgn(s-maxs)>0)
 61                 {
 62                     k++;
 63                     maxs=s;
 64                 }
 65                 else
 66                 break;
 67             }
 68             vdp[i*(n-2)+j-2].first=maxs;
 69             if(k<j)
 70             vdp[i*(n-2)+j-2].second=k-1;
 71             else
 72             vdp[i*(n-2)+j-2].second=j-1;            
 73         }
 74     }
 75     for(int l=1;l2;l++)
 76     {
 77         for(int i=0;i2;i++)
 78         {
 79             double maxs=0.0;
 80             int kk=0;
 81             for(int j=i+l+2;j)
 82             {
 83                 for(int jj=i+l+1;jj)
 84                 if(sgn(maxs-vdp[(l-1)*(n-2)*(n-2)+i*(n-2)+jj-2].first-area(jj,j)-area(j,i)-area(i,jj))<0)
 85                 {
 86                     maxs=vdp[(l-1)*(n-2)*(n-2)+i*(n-2)+jj-2].first+area(jj,j)+area(j,i)+area(i,jj);
 87                     kk=jj;
 88                 }
 89                 vdp[l*(n-2)*(n-2)+i*(n-2)+j-2].first=maxs;
 90                 vdp[l*(n-2)*(n-2)+i*(n-2)+j-2].second=kk;
 91             }
 92         }
 93     }
 94     double maxs=0.0;
 95     int ii=0,jj=0;
 96     for(int i=0;i<(n-m+1);i++)
 97     {
 98         for(int j=i+m-3;j2;j++)
 99         {
100             if(sgn(maxs-vdp[(m-3)*(n-2)*(n-2)+i*(n-2)+j].first)<0)
101             {
102                 maxs=vdp[(m-3)*(n-2)*(n-2)+i*(n-2)+j].first;
103                 ii=i;
104                 jj=j;
105             }
106         }
107     }
108     vector<int> vr;
109     vr.emplace_back(v[ii].id);
110     vr.emplace_back(v[jj+2].id);
111     for(int i=m-3;i>=0;i--)
112     {
113         vr.emplace_back(v[vdp[i*(n-2)*(n-2)+ii*(n-2)+jj].second].id);
114         jj=vdp[i*(n-2)*(n-2)+ii*(n-2)+jj].second-2;
115     }
116     sort(vr.begin(),vr.end(),greater<int>());
117     printf("%d",vr[0]);
118     for(int i=1;i)
119     printf(" %d",vr[i]);
120     return 0;
121 }