基于水平序的Andrew凸包算法 最详细的图解（多图预警）

给出凸包的定义：

简要说一下思路：

首先将所有点按照x从小到大（x同则y从小到大）排序
把p1,p2放入凸包，从p3开始，当新点在凸包‘前进’方向的左边时继续，否则依次删除最近加入凸包的点，直到新点在左边
输入不能有重复点，不希望凸包边上有点可疑将<=改为<

 

下面请根据程序，测试数据与输出 及步骤图学习，跟着模拟一遍就会非常明白了

#include
using namespace std;
const double eps=1e-10;                                                              
struct Point{                                                                        
    double x,y;                                                                     
    int id;
    Point(double x=0,double y=0,int id=0):x(x),y(y),id(id){}
};
typedef Point Vector;

bool operator< (const Point&a,Point &b){
	return a.x0)?1:-1;}
bool operator == (const Vector A,const Vector B){
    return dcmp(A.x-B.x)==0 && dcmp(A.y-B.y)==0;
}
double Dot(Vector A,Vector B){return A.x*B.x+A.y*B.y;}  
double Length(Vector A){return sqrt(Dot(A,A));}         
double Cross(Vector A,Vector B){return A.x*B.y-B.x*A.y;}

int ConvexHell(Point p[],int n,Point ch[]) {
	sort(p,p+n);
	int m=0;
	for(int i=0;i1 && Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0){           //下凸包 
			printf("update: segment %d TO %d destination: %d\n",m-2,m-1,i);
			m--;
		}
		ch[m++]=p[i];
		printf("ch[ %d ]= %d\n",m-1,i);
	}
	int k=m;
	for(int i=n-2;i>=0;i--){
		while(m>k && Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0){          //上凸包 
			printf("update: segment %d TO %d destination: %d\n",m-2,m-1,i);
			m--;
		}
		ch[m++]=p[i];
		printf("ch[ %d ]= %d\n",m-1,i);
	}
	if(n>1)m--;
	return m;
}
int main(){
	int n;
	scanf("%d",&n);
	Point p[20],ch[20];
	for(int i=0;i

/*测试数据及输出
8
3 0
4 8
5 5
6 3
6 7
8 4
9 2
10 7
ch[ 0 ]= 0
ch[ 1 ]= 1
update: segment 0 TO 1 destination: 2
ch[ 1 ]= 2
update: segment 0 TO 1 destination: 3
ch[ 1 ]= 3
ch[ 2 ]= 4
update: segment 1 TO 2 destination: 5
update: segment 0 TO 1 destination: 5
ch[ 1 ]= 5
update: segment 0 TO 1 destination: 6
ch[ 1 ]= 6
ch[ 2 ]= 7
ch[ 3 ]= 6
update: segment 2 TO 3 destination: 5
ch[ 3 ]= 5
update: segment 2 TO 3 destination: 4
ch[ 3 ]= 4
ch[ 4 ]= 3
update: segment 3 TO 4 destination: 2
ch[ 4 ]= 2
update: segment 3 TO 4 destination: 1
update: segment 2 TO 3 destination: 1
ch[ 3 ]= 1
ch[ 4 ]= 0
Final Ans:
3 0,9 2,10 7,4 8,
*/