poj 1873（枚举所有的状态+凸包）

#include <iostream>
#include <cstdio>
#include <cstring>
#include <math.h>
#include <algorithm>
#include <stdlib.h>
using namespace std;
const double eps = 1e-8;
const int N = 20;
struct Point
{
    double x,y,value,len;
} p[N],q[N];
Point Stack[N];
int n;
double cross(Point a,Point b,Point c)
{
    return (a.x-c.x)*(b.y-c.y)-(a.y-c.y)*(b.x-c.x);
}
double dis(Point a,Point b)
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
Point po; //大坑
int cmp(Point a,Point b)
{
    if(cross(a,b,po)>0) return 1;
    if(cross(a,b,po)==0&&dis(b,po)-dis(a,po)>eps) return 1;
    return 0;
}
double Graham(Point p[],int n)
{
    if(n==0||n==1) return 0;
    if(n==2) return 2*dis(p[0],p[1]);
    int k =0;
    for(int i=0; i<n; i++)
    {
        if(p[k].y>p[i].y||((p[k].y==p[i].y)&&(p[k].x>p[i].x))) k=i;
    }
    swap(p[0],p[k]);
    po = p[0];  ///嗯，这里
    int top=2;
    sort(p+1,p+n,cmp);
    Stack[0]=p[0];
    Stack[1]=p[1];
    Stack[2]=p[2];
    for(int i=3; i<n; i++)
    {
        while(top>=1&&cross(p[i],Stack[top],Stack[top-1])>=0)
        {
            top--;
        }
        Stack[++top]=p[i];
    }
    double ans = 0;
    for(int i=1; i<=top; i++)
    {
        ans+=dis(Stack[i],Stack[i-1]);
    }
    ans+=dis(Stack[top],Stack[0]);
    return ans;
}
int main()
{
    int _count = 1;
    while(scanf("%d",&n)!=EOF&&n)
    {
        for(int i=0; i<n; i++)
        {
            scanf("%lf%lf%lf%lf",&p[i].x,&p[i].y,&p[i].value,&p[i].len);
        }
        int t = (1<<n)-1;
        int cnt[N],id[N];///保存要砍掉的树的编号
        double save=0;
        int num=999999999; ///要砍掉的数的数目
        double mi=999999999; ///保存要砍掉的树的最小价值
        for(int i=1; i<t; i++) ///1代表砍掉这棵树,0代表不砍,至少要砍一棵树
        {
            double value=0,len=0;
            int k=0,k1=0;
            for(int j=0; j<n; j++)
            {
                if((i>>j)&1)  ///移位操作,表示当前第j+1棵树要砍掉
                {
                    cnt[k++] = j;
                    value+=p[j].value;
                    len+=p[j].len;
                }
                else q[k1++] = p[j];
            }
            double L = Graham(q,k1);
            if(len-L>eps)   ///如果能够组成的篱笆长度大于凸包周长
            {
                if(mi-value>eps||(fabs(value-mi)<eps&&k<num))
                {
                    mi = value;
                    for(int j=0; j<k; j++) id[j]=cnt[j];
                    num = k;
                    save = len - L;
                }
            }
        }
        printf("Forest %d\nCut these trees: ",_count++);
        for(int i=0; i<num; i++)
        {
            printf("%d ",id[i]+1);
        }
        printf("\nExtra wood: %.2lf\n\n",save);
    }
    return 0;
}