优化算法：二进制搜索与凸包求解在森林砍伐问题中的应用-优快云博客

题目链接：1873 -- The Fortified Forest (poj.org)
二进制枚举+凸包
用S表示砍掉的树的集合，一棵树对应二进制的一位 S=5 （101）砍掉第一棵树和第三棵树

/*
floor();//向下取整  ceil();//向上取整  round();//四舍五入
判等时不要直接使用=
*/
//#include<bits/stdc++.h>
#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<cstdio>
#include<vector>
#include<climits>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
using namespace std;
const double pi = acos(-1.0);
const double inf = 1e100;
/*
误差判断
*/
const double eps = 1e-8;
int dcmp(double x, double y){
    if(fabs(x - y) < eps)
        return 0;
    if(x > y)
        return 1;
    return -1;
}
int sgn(double d){
    if(fabs(d) < eps)
        return 0;
    if(d > 0)
        return 1;
    return -1;
}
/*---------------------------------------------------------------------------------------*/
//点
struct Point{
	double x, y,z;
    Point(double x = 0, double y = 0, double z = 0):x(x),y(y),z(z){}
};
//向量
typedef Point Vector;
//运算
Vector operator + (Vector A, Vector B){
    return Vector(A.x+B.x, A.y+B.y,A.z+B.z);
}
Vector operator - (Vector A, Vector B){
    return Vector(A.x-B.x, A.y-B.y,A.z-B.z);
}
Vector operator * (Vector A, double p){
    return Vector(A.x*p, A.y*p,A.z*p);
}
Vector operator / (Vector A, double p){
    return Vector(A.x/p, A.y/p,A.z/p);
}
bool operator == (Point A,Point B){
	return sgn(A.x-B.x)==0&&sgn(A.y-B.y)==0;
}
bool P_cmp1(Point a,Point b){
    if(a.x==b.x)
    	return a.y<b.y;
    return a.x<b.x;
}
//向量数量积  向量α在向量β的投影于向量β的长度乘积（带方向）
double Dot(Vector A, Vector B){
    return A.x*B.x + A.y*B.y + A.z*B.z;
}
//向量向量积  向量α与β所张成的平行四边形的有向面积
double Cross(Vector A, Vector B){
    return A.x*B.y-A.y*B.x;
}
//取模
double Length(Vector A){
    return sqrt(Dot(A, A));
}
//计算夹角
double Angle(Vector A, Vector B){
    return acos(Dot(A, B) / Length(A) / Length(B));
}
//向量旋转
Vector Rotate(Vector A, double rad){//rad为弧度 且为逆时针旋转的角
    return Vector(A.x*cos(rad)-A.y*sin(rad), A.x*sin(rad)+A.y*cos(rad));
}
Vector Normal(Vector A){//向量A左转90°的单位法向量
    double L = Length(A);
    return Vector(-A.y/L, A.x/L);
}
//三角形外接圆圆心
Point Excenter(Point a, Point b, Point c){
    double a1 = b.x - a.x;
    double b1 = b.y - a.y;
    double c1 = (a1*a1 + b1*b1)/2;
    double a2 = c.x - a.x;
    double b2 = c.y - a.y;
    double c2 = (a2*a2 + b2*b2)/2;
    double d = a1*b2 - a2*b1;
    return Point(a.x + (c1*b2 - c2*b1)/d, a.y + (a1*c2 - a2*c1)/d);
}
//两点距离
double dis(Point a,Point b){
	return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)+(a.z-b.z)*(a.z-b.z));
}
/*------------------------------------------------------------------------------------------------*/
//线
struct Line{//直线定义
    Point v, p;
    //Line(Point v, Point p):v(v), p(p){}
    Point point(double t){//返回点P = v + (p - v)*t
        return v + (p - v)*t;
    }
};
//判断点是否在线上
bool PointOnline(Line A,Point a){
	Vector v1(A.p.x-a.x,A.p.y-a.y,A.p.z-a.z);
	Vector v2(A.v.x-a.x,A.v.y-a.y,A.v.z-a.z);
	return !(Cross(v1,v2));
}
//计算两直线交点 必须保证直线相交，否则将会出现除以零的情况
Point GetLineIntersection(Point a1, Point a2, Point b1, Point b2){
	Vector v = a1-a2;
	Vector w = b1-b2;
	Vector u = a1-b1;
    double t = Cross(w, u)/Cross(v, w);
    return a1+v*t;
}
//点P到直线AB距离公式
double DistanceToLine(Point P, Point A, Point B){
    Vector v1 = B-A, v2 = P-A;
    return fabs(Cross(v1, v2)/Length(v1));
}
//点P到线段AB距离公式
double DistanceToSegment(Point P, Point A, Point B){
    if(A.x == B.x&&A.y == B.y&&A.z == B.z)
        return Length(P-A);
    Vector v1 = B-A, v2 = P-A, v3 = P-B;
    if(dcmp(Dot(v1, v2),0) < 0)
        return Length(v2);
    if(dcmp(Dot(v1, v3),0) > 0)
        return Length(v3);
    return DistanceToLine(P, A, B);
}
//点P在直线AB上的投影点
Point GetLineProjection(Point P, Point A, Point B){
    Vector v = B-A;
    return A+v*(Dot(v, P-A)/Dot(v, v));
}
//判断点是否在线段端上
bool OnSegment(Point p, Point a1, Point a2){
    return dcmp(Cross(a1-p, a2-p),0) == 0 && dcmp(Dot(a1-p, a2-p),0) <= 0;
}
//判断线段相交
bool SegmentProperIntersection(Point a1, Point a2, Point b1, Point b2){
    double c1 = Cross(a2-a1, b1-a1), c2 = Cross(a2-a1, b2-a1);
    double c3 = Cross(b2-b1, a1-b1), c4 = Cross(b2-b1, a2-b1);
    //if判断控制是否允许线段在端点处相交，根据需要添加
    if(!sgn(c1) || !sgn(c2) || !sgn(c3) || !sgn(c4)){
        bool f1 = OnSegment(b1, a1, a2);
        bool f2 = OnSegment(b2, a1, a2);
        bool f3 = OnSegment(a1, b1, b2);
        bool f4 = OnSegment(a2, b1, b2);
        bool f = (f1|f2|f3|f4);
        return f;
    }
    return (sgn(c1)*sgn(c2) < 0 && sgn(c3)*sgn(c4) < 0);
}
//点c是否在线段ab的左侧 左1 线上0 右-1
int ToLeftTest(Point a, Point b, Point c){
    if( Cross(b - a, c - b) > 0) return 1;
    else if( Cross(b - a, c - b) == 0) return 0;
    else return -1;
}
/*------------------------------------------------------------------------*/
//多边形
//求凸包 多边形p  凸包ch
int ConvexHull(Point *p,int n,Point *ch){
	n=unique(p,p+n)-p;//去重
	sort(p,p+n,P_cmp1);
	int v=0;
	for(int i=0;i<n;i++){
		while(v>1&&Cross(ch[v-1]-ch[v-2],p[i]-ch[v-1])<=0)
			v--;
		ch[v++]=p[i];
	}
	int j=v;
	for(int i=n-2;i>=0;i--){
		while(v>j&&Cross(ch[v-1]-ch[v-2],p[i]-ch[v-1])<=0)
			v--;
		ch[v++]=p[i];
	}
	if(n>1) v--;
	return v;
}
//求凸包周长
double CofCH(Point *ch,int m){
	double C=0;
	for(int i=0;i<m;i++)
		C+=dis(ch[i],ch[(i+1)%m]);
	return C;
}
//多边形有向面积
double PolygonArea(Point* p, int n){//p为端点集合，n为端点个数
    double s = 0;
    for(int i = 1; i < n-1; ++i)
        s += Cross(p[i]-p[0], p[i+1]-p[0]);
    return s;
}
//判断点是否在多边形内，若点在多边形内返回1，在多边形外部返回0，在多边形上返回-1
int isPointInPolygon(Point p, vector<Point> poly){
    int wn = 0;
    int n = poly.size();
    for(int i = 0; i < n; ++i){
        if(OnSegment(p, poly[i], poly[(i+1)%n])) return -1;
        int k = sgn(Cross(poly[(i+1)%n] - poly[i], p - poly[i]));
        int d1 = sgn(poly[i].y - p.y);
        int d2 = sgn(poly[(i+1)%n].y - p.y);
        if(k > 0 && d1 <= 0 && d2 > 0) wn++;
        if(k < 0 && d2 <= 0 && d1 > 0) wn--;
    }
    if(wn != 0)
        return 1;
    return 0;
}
/************************************************************************************/
Point Tree[20],p[20],ch[20];
int k=1;
double v[20],l[20];
void isok(int n)
{
	for(int i=0;i<n;i++)
		cin>>Tree[i].x>>Tree[i].y>>v[i]>>l[i];
	double min_val=1e8; //砍树最小价值和
    double min_len;     //当前剩余木材数
    int min_num=1e8;    //砍树最小数目
    int min_state;      //最佳状态
	//二份枚举  枚举所有可能   1砍掉 
	for(int S=1;S<(1<<n);S++)
	{
		double cut_val=0;//当前砍树总价值
        double cut_len=0;//木材总长度
        int cut_num=0;   //砍树总颗数
		int ucut_num=0;  //未被砍下的树	
		for(int i=0;i<n;i++)
		{
			if(S&(1<<i))
			{
				cut_val+=v[i];
				cut_len+=l[i];
				cut_num++;	
			}
			else
				p[ucut_num++]=Tree[i];
		}
		//构建凸包
		int m=ConvexHull(p,ucut_num,ch);
		//求凸包周长
		double C=CofCH(ch,m);
  		if(cut_len>=C)//砍下的树比凸包周长长
		{
			if(min_val>cut_val||(min_val==cut_val&&min_num>cut_num))
            {
                min_val=cut_val;
                min_num=cut_num;
                min_state=S;
                min_len=cut_len-C;
//                cout<<S<<endl;
            }
		}
	}
//	cout<<min_state<<endl;
	printf("Forest %d\nCut these trees:",k);
	k++;
//	cout<<endl<<min_num<<endl;
    for(int i=0;i<n;i++)
		if(min_state&(1<<i))
        	printf(" %d",i+1);
    printf("\nExtra wood: %.2lf\n",min_len);
    cout<<endl;
}
int main()
{
	int n;
	while(1)
	{
		cin>>n;
		if(n==0)
			return 0;
		isok(n);
	}
}
POJ：1873
