题意：是有n棵树，每棵的坐标，价值和长度已知，要砍掉若干根，用他们围住其他树，问损失价值最小的情况下又要长度足够围住其他树，砍掉哪些树。。

思路：先求要砍掉的哪些树，在求剩下的树求凸包，在判是否可行。（枚举+凸包）

// Time 407ms; Memory 200K

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#define inf 200000
using namespace std;bool vis[15],tvis[15];
int n,m,minv,tn;typedef struct point
{double x,y;int v;double l;int z;point(double xx=0,double yy=0,int vv=0,double ll=0,int zz=0):x(xx),y(yy),v(vv),l(ll),z(zz){}
}vector;point p[15],ch[15];bool operator < (point a,point b)
{return a.x<b.x || (a.x==b.x && a.y<b.y);
}
vector operator - (point a,point b)
{return vector(a.x-b.x,a.y-b.y);
}
double cross(vector a,vector b)
{return a.x*b.y-a.y*b.x;
}
double len(vector a)
{return sqrt(a.x*a.x+a.y*a.y);
}int graph()
{int k,i;m=0;for(i=0;i<n;i++) if(!vis[i]){while(m>1 && cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0) m--;ch[m++]=p[i];}k=m;for(i=n-2;i>=0;i--) if(!vis[i]){while(m>k && cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0) m--;ch[m++]=p[i];}if(n>1) m--;double s1=0,s2=0;for(i=0;i<m;i++) s1+=len(ch[i]-ch[(i+1)%m]);for(i=0;i<n;i++) if(vis[i]) s2+=p[i].l;return s1<=s2;
}
void dfs(int d)
{int i,mv=0,mn=0;if(d==n){if(graph()){for(i=0;i<n;i++) if(vis[i]) {mv+=p[i].v;mn++;}if(mv<minv || (mv==minv && tn>mn)) {for(i=0;i<n;i++) tvis[i]=vis[i];minv=mv;tn=mn;}}return;}dfs(d+1);vis[d]=1;dfs(d+1);vis[d]=0;
}
int main()
{int i,j,t=0,a[15];while(scanf("%d",&n)!=EOF && n){for(i=0;i<n;i++){scanf("%lf%lf%d%lf",&p[i].x,&p[i].y,&p[i].v,&p[i].l);p[i].z=i;}sort(p,p+n);minv=inf;tn=inf;memset(vis,0,sizeof(vis));dfs(0);if(t++) printf("\n");printf("Forest %d\n",t);printf("Cut these trees:");j=0;for(i=0;i<n;i++) if(tvis[i]) a[j++]=p[i].z+1;sort(a,a+j);for(i=0;i<j;i++) printf(" %d",a[i]);double s1=0,s2=0;for(i=0;i<n;i++) vis[i]=tvis[i];graph();for(i=0;i<m;i++) s1+=len(ch[i]-ch[(i+1)%m]);for(i=0;i<n;i++) if(vis[i]) s2+=p[i].l;printf("\nExtra wood: %.2lf\n",s2-s1);}return 0;
}