poj 1873 枚举+凸包

题意：给出一些树的位置，价值，长度，现要求先砍一些树制成一定长度的篱笆将剩余的树围起来，求要砍树的最小总价值。

因为树的个数最多为15个很容易想到用二进制数表示树的状态进行遍历。

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
#define MAX_INT 123456789
#define esp 10e-8
struct point 
{
    int x,y;
};
point vertex[16];
int res[16],Choice[16],noChoice[16],minLeft[16];
double length[16],wight[16];
int cmp(const void * a,const void * b)
{
    point p1,p2;
    p1=vertex[*(int*)a]; p2=vertex[*(int *)b];
    if(p1.x==p2.x)
        return p1.y-p2.y;
    return p1.x-p2.x;
}
int cross(point p0,point p1,point p2)
{
    return (p2.y-p0.y)*(p1.x-p0.x)-(p1.y-p0.y)*(p2.x-p0.x);
}
int dist(point p1,point p2)
{
    return (p1.y-p2.y)*(p1.y-p2.y)+(p1.x-p2.x)*(p1.x-p2.x);
}
int cmp1(const void* a,const void * b)
{
    return *(int *)a-*(int *)b;
}
int graham(int n,int m)
{
    int i,len,top=1;
    if(n<2) 
        return 0;
    qsort(Choice,n,sizeof(Choice[0]),cmp);

    res[0]=Choice[0]; res[1]=Choice[1];

    for(i=2;i<n;i++)
    {
        while(top && cross(vertex[res[top-1]],vertex[res[top]],vertex[Choice[i]])<=0)
            top--;
        res[++top]=Choice[i];
    }
    len=top; res[++top]=Choice[n-2];
    for(i=n-3;i>=0;i--)
    {
        while(top!=len && cross(vertex[res[top-1]],vertex[res[top]],vertex[Choice[i]])<=0)
            top--;
        res[++top]=Choice[i];
    }
    return top;
}
int print(int count,int len,double leaves)
{
    int i;
    qsort(minLeft,len,sizeof(int),cmp1);

    printf("Forest %d\n",count);
    
    printf("Cut these trees:");
    for(i=0;i<len;i++)
        printf(" %d",minLeft[i]+1);
    printf("\n");
    printf("Extra wood: %.2lf\n\n",leaves);
    return 0;
}
int main()
{
    int i,j,k,n,len,len1,len2,top,count=0;
    double cutLen,noCutLen,leaves,value, cur_min;
    while(scanf("%d",&n) && n)
    {
        count++;
        for(i=0;i<n;i++)
            scanf("%d%d%lf%lf",&vertex[i].x,&vertex[i].y,&wight[i],&length[i]);
        k=(1<<n);  cur_min=MAX_INT; len=0;
        for(i=k-1;i>=0;i--)
        {
            len1=len2=0; 
            memset(Choice,0,sizeof(Choice));
            //选择要砍的树和不砍的树
            for(j=0;j<n;j++)
            {
                if(i&(1<<j)) 
                    Choice[len1++]=j;
                else
                    noChoice[len2++]=j;
            }
            cutLen=noCutLen=value=0;
            //砍的树的总长度和总价值
            for(j=0;j<len2;j++)  
            {
                cutLen+=length[noChoice[j]];
                value+=wight[noChoice[j]];
            }
            //计算凸包和周长
            top=graham(len1,len2);

            for(j=0;j<top;j++) 
            {
                noCutLen+=sqrt((double)dist(vertex[res[j]],vertex[res[j+1]]));
            }
            //寻找在砍的树的总长度大于凸包周长的情况下的最小价值
            if(cutLen>=noCutLen)
            {
                if(value<cur_min || (abs(value-cur_min)<=esp && len2<len))
                {
                    cur_min=value; len=len2;

                    leaves=cutLen-noCutLen;

                    for(j=0;j<len2;j++)
                    {
                        minLeft[j]=noChoice[j];
                    }
                }
            }
        }

        print(count,len,leaves);
    }
    return 0;
}