凸包模板 POJ1873

Posted 2020-08-09 blues

  1 // 凸包模板 POJ1873
  2 // n=15所以可以按位枚举求凸包，再记录数据
  3 
  4 #include <iostream>
  5 #include <cstdio>
  6 #include <cstdlib>
  7 #include <algorithm>
  8 #include <vector>
  9 #include <math.h>
 10 using namespace std;
 11 #define LL long long
 12 typedef pair<int,int> pii;
 13 const double inf = 0x3f3f3f3f;
 14 const LL MOD =100000000LL;
 15 const int N =110;
 16 #define clc(a,b) memset(a,b,sizeof(a))
 17 const double eps = 1e-8;
 18 void fre() {freopen("in.txt","r",stdin);}
 19 void freout() {freopen("out.txt","w",stdout);}
 20 inline int read() {int x=0,f=1;char ch=getchar();while(ch>‘9‘||ch<‘0‘) {if(ch==‘-‘) f=-1; ch=getchar();}while(ch>=‘0‘&&ch<=‘9‘) {x=x*10+ch-‘0‘;ch=getchar();}return x*f;}
 21 
 22 int sgn(double x){
 23     if(fabs(x) < eps)return 0;
 24     if(x < 0)return -1;
 25     else return 1;
 26 }
 27 
 28 struct Point{
 29     double x,y;
 30     Point(){}
 31     Point(double _x,double _y){
 32         x = _x;y = _y;
 33     }
 34     Point operator -(const Point &b)const{
 35         return Point(x - b.x,y - b.y);
 36     }
 37     double operator ^(const Point &b)const{
 38         return x*b.y - y*b.x;
 39     }
 40     double operator *(const Point &b)const{
 41         return x*b.x + y*b.y;
 42     }
 43     friend bool operator<(const Point &a,const Point &b){
 44         if(fabs(a.y-b.y)<eps) return a.x<b.x;
 45         return a.y<b.y;
 46     }
 47     friend double dis2(Point a){
 48         return a.x*a.x+a.y*a.y;
 49     }
 50 };
 51 
 52 double dis(Point a,Point b){
 53     return sqrt(dis2(a-b));
 54 }
 55 double mult(Point a,Point b,Point o){
 56     return (a.x-o.x)*(b.y-o.y)>=(b.x-o.x)*(a.y-o.y);
 57 }
 58 
 59 Point P[16];
 60 double v[16],l[16];
 61 
 62 double graham(Point p[],int n,Point q[]){
 63      int top=1;
 64      sort(p,p+n);
 65      if(n==0) return 0;
 66      q[0]=p[0];
 67      if(n==1) return 0;
 68      q[1]=p[1];
 69      if(n==2) return dis(p[0],p[1])*2;
 70      q[2]=p[2];
 71      for(int i=2;i<n;i++){
 72          while(top&&(mult(p[i],q[top],q[top-1]))) top--;
 73          q[++top]=p[i];
 74      }
 75      int len=top;
 76      q[++top]=p[n-2];
 77      for(int i=n-3;i>=0;i--){
 78          while(top!=len&&(mult(p[i],q[top],q[top-1]))) top--;
 79          q[++top]=p[i];
 80      }
 81      double c=dis(q[0],q[top-1]);
 82      for(int i=0;i<top-1;i++)
 83         c+=dis(q[i],q[i+1]);
 84     return c;
 85 }
 86 
 87 int n;
 88 int b[16],a[16];
 89 Point p[16],ch[16];
 90 int main(){
 91     int cas=1;
 92     while(~scanf("%d",&n),n){
 93         for(int i=0;i<n;i++){
 94             double x,y;
 95             scanf("%lf%lf%lf%lf",&x,&y,&v[i],&l[i]);
 96             P[i]=Point(x,y);
 97         }
 98         int k,cnt;
 99         double ans=0;
100         int minn_cnt=11111;
101         double sum,len,minn=inf;
102         for(int st=0;st<(1<<n);st++){
103             sum=0,len=0,k=0,cnt=0;
104 
105             for(int i=0;i<n;i++){
106                if(st&(1<<i)){
107                    sum+=v[i];
108                    len+=l[i];
109                    b[++cnt]=i+1;
110                }
111                else{
112                   p[k++]=P[i];
113                }
114             }
115             double lenn=graham(p,k,ch);
116             if(lenn>len) continue;
117             if(sum<minn||(sum==minn&&cnt<minn_cnt)){
118                 minn=sum;
119                 minn_cnt=cnt;
120                 ans=len-lenn;
121                 for(int i=1;i<=cnt;i++){
122                     a[i]=b[i];
123                 }
124             }
125         }
126         printf("Forest %d\n",cas++);
127         printf("Cut these trees:");
128         for(int i=1;i<=minn_cnt;i++){
129             printf(" %d",a[i]);
130         }
131         printf("\n");
132         printf("Extra wood: ");
133         printf("%.2f\n",ans);
134         printf("\n");
135     }
136     return 0;
137 }