poj1873 The Fortified Forest 凸包+枚举 水题
/* poj1873 The Fortified Forest 凸包+枚举 水题 用小树林的木头给小树林围一个围墙 每棵树都有价值 求消耗价值最低的做法,输出被砍伐的树的编号和剩余的木料 若砍伐价值相同,则取砍伐数小的方案。 */ #include<stdio.h> #include<math.h> #include <algorithm> #include <vector> using namespace std; const double eps = 1e-8; struct point { double x,y; }; struct exinfo { int v,l; }info[20]; int n; point dian[20],zhan[20]; ////////////////////////////////////////////////// point *mo_dian; double mo_distance(point p1,point p2) { return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y)); } double mo_xmult(point p2,point p0,point p1)//p1在p2左返回负,在右边返回正 { return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y); } bool mo_ee(double x,double y) { double ret=x-y; if(ret<0) ret=-ret; if(ret<eps) return 1; return 0; } bool mo_gg(double x,double y) { return x > y + eps;} // x > y bool mo_ll(double x,double y) { return x < y - eps;} // x < y bool mo_ge(double x,double y) { return x > y - eps;} // x >= y bool mo_le(double x,double y) { return x < y + eps;} // x <= y bool mo_cmp(point a,point b) // 第一次排序 { if( mo_ee(a.y ,b.y ) ) return mo_ll(a.x, b.x); return mo_ll(a.y,b.y); } bool mo_cmp1(point a,point b) // 第二次排序 { double len = mo_xmult(b,mo_dian[0],a); if( mo_ee(len,0.0) ) return mo_ll(mo_distance(mo_dian[0],a),mo_distance(mo_dian[0],b)); return mo_gg(len,0.0); } int mo_graham(int n,point *dian,point *stk) { int i,top=1; mo_dian=dian; sort(mo_dian,mo_dian+n,mo_cmp); sort(mo_dian+1,mo_dian+n,mo_cmp1); stk[0]=mo_dian[0]; stk[1]=mo_dian[1]; for(i=2;i<n;++i) { while(top>0&&mo_xmult(mo_dian[i],stk[top-1],stk[top])<=0) --top; stk[++top]=mo_dian[i]; } return top+1; } //////// void getpoint(int tree,point *dian,point *work,vector<int> &cut,int &n_cut,int &n_sheng,int &cutv,int &cutl) { cut.clear(); cutv=0; cutl=n_sheng=n_cut=0; int i; for(i=0;i<n;++i) { if((1<<i)&tree)//cut { cut.push_back(i); n_cut++; cutv+=info[i].v; cutl+=info[i].l; }else { work[n_sheng++]=dian[i]; } } } double zhou(point *dian,int n) { int i; double ret=0; for(i=0;i<n;++i) { ret+=sqrt( (dian[(i+1)%n].x-dian[i].x)*(dian[(i+1)%n].x-dian[i].x) +(dian[(i+1)%n].y-dian[i].y)*(dian[(i+1)%n].y-dian[i].y) ); } return ret; } int main() { int i,ncase=1; while(scanf("%d",&n),n) { for(i=0;i<n;++i) { scanf("%lf%lf%d%d",&dian[i].x,&dian[i].y,&info[i].v,&info[i].l); } int maxofi=(1<<n)-1; int cutvalue=999999999; double l_sheng; int cutn; vector<int> cut; vector<int> tempcut; point work[20]; for(i=0;i<maxofi;++i) { int tempcutn,shengyun; int tempcutv,cutl; int ret; double zhouchang; getpoint(i,dian,work,tempcut,tempcutn,shengyun,tempcutv,cutl); if(shengyun==1) { zhouchang=0; }else if(shengyun==2) { zhouchang=mo_distance(work[0],work[1])*2; }else { ret=mo_graham(shengyun,work,zhan); zhouchang=zhou(zhan,ret); } if(mo_ge(cutl,zhouchang)) { if(tempcutv<cutvalue) { cutvalue=tempcutv; l_sheng=cutl-zhouchang; cut=tempcut; cutn=tempcutn; }else if(tempcutv==cutvalue&&tempcutn<cutn) { cutvalue=tempcutv; l_sheng=cutl-zhouchang; cut=tempcut; cutn=tempcutn; } } } if(ncase!=1) printf("\n"); printf("Forest %d\n",ncase++); printf("Cut these trees:"); int len=cut.size(); for(i=0;i<len;++i) { printf(" %d",cut[i]+1); } printf("\n"); printf("Extra wood: %.2lf\n",l_sheng); } return 0; }