hdu 3685 计算几何（好）

判断重心在每条凸包边上的的垂足是否在凸包边上，计算几何一般思路很清晰，就是实现起来有点烦，这里错一点，那里错一点，所以，以后决定要把几何题放最后做了

第一种方法：求出垂足再判断

#include<stdio.h>

#include<string.h>

#include<math.h>

#include<algorithm>

using namespace std;

const double eps = 1e-8;

struct point {

	double x,y;

	point operator - (const point& t) const {

        point tmp;

        tmp.x = x - t.x;

        tmp.y = y - t.y;

        return tmp;

    }

    point operator + (const point& t) const {

        point tmp;

        tmp.x = x + t.x;

        tmp.y = y + t.y;

        return tmp;

    }

	bool operator == (const point& t)const {

		return fabs(x-t.x) < eps && fabs(y-t.y) < eps;

	}

}p[100010];

struct line{

	point a,b;

};

int top;

bool cmpxy(point a,point b){

	if(a.y==b.y)

		return a.x<b.x;

	return a.y<b.y;

}

double cross(point a,point b,point c){

	return (b.x-a.x)*(c.y-a.y)-(c.x-a.x)*(b.y-a.y);

}

void tubao(point *p, int n) {							

	if ( n < 3 ) 

		return;

	int i, m=0;top=1;

	sort(p, p+n, cmpxy);	

	for (i=n; i < 2*n-1; i++)

		p[i] = p[2*n-2-i];

	for (i=2; i < 2*n-1; i++) {

		while ( top > m && cross(p[top], p[i], p[top-1]) < eps ) 

			top--;

		p[++top] = p[i];

		if ( i == n-1 )	m = top;

	}

}

point gravi(point* p, int n) {							

	int i;

	double A=0, a;

	point t;

	t.x = t.y = 0;

	p[n] = p[0];

	for (i=0; i < n; i++) {

		a = p[i].x*p[i+1].y - p[i+1].x*p[i].y;

		t.x += (p[i].x + p[i+1].x) * a;

		t.y += (p[i].y + p[i+1].y) * a;

		A += a;

	}

	t.x /= A*3;

	t.y /= A*3;

	return t;

}

point intersection(point u1,point u2,point v1,point v2){

	point ret=u1;

	double t=((u1.x-v1.x)*(v1.y-v2.y)-(u1.y-v1.y)*(v1.x-v2.x))

			/((u1.x-u2.x)*(v1.y-v2.y)-(u1.y-u2.y)*(v1.x-v2.x));

	ret.x+=(u2.x-u1.x)*t;

	ret.y+=(u2.y-u1.y)*t;

	return ret;

}

point ptoline(point p,point l1,point l2){

	point t=p;

	t.x+=l1.y-l2.y,t.y+=l2.x-l1.x;

	return intersection(p,t,l1,l2);

}

bool dot_onseg(point p, point s, point e) {		

    if ( p == s || p == e )	

        return false;

    return cross(p,s,e) < eps  && 

        (p.x-s.x)*(p.x-e.x)<eps && (p.y-s.y)*(p.y-e.y)<eps;

}

int main()

{

	int t,i,j,n;

	scanf("%d",&t);

	while(t--)

	{

		scanf("%d",&n);

		for(i=0;i<n;i++)

			scanf("%lf%lf",&p[i].x,&p[i].y);

		point cen=gravi(p,n); 

		tubao(p,n);

		line seg;point m; 

		int count=0;

		for(i=0;i<top;i++)

		{

			m=ptoline(cen,p[i],p[i+1]);

            if(dot_onseg(m,p[i],p[i+1]))

				count++;

		}

		printf("%d\n",count);

	}

	return 0;

}

第二种方法：直接利用点积的性质判断，从重心像凸包某条边上的两个点连线，两个角都应为锐角

View Code

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
using namespace std;
const double eps = 1e-8;
struct point {
    double x,y;
}p[100010];
struct line{
    point a,b;
};int top;
bool cmpxy(point a,point b){
    if(fabs(a.y-b.y)>eps)
         return a.y<b.y;
   return a.x<b.x;
}
double cross(point a,point b,point c){
    return (b.x-a.x)*(c.y-a.y)-(c.x-a.x)*(b.y-a.y);
}
void tubao(point *p, int n) {                            
    if ( n < 3 ) 
        return;
    int i, m=0;top=1;
    sort(p, p+n, cmpxy);    
    for (i=n; i < 2*n-1; i++)  p[i] = p[2*n-2-i];
    for (i=2; i < 2*n-1; i++){
        while ( top > m && cross(p[top], p[i], p[top-1]) < eps ) 
            top--;
        p[++top] = p[i];
        if ( i == n-1 )    m = top;
    }
}
point gravi(point* p, int n) {                            
    int i;
    double A=0, a;
    point t;t.x = t.y = 0;
    p[n] = p[0];
    for (i=0; i < n; i++) {
        a = p[i].x*p[i+1].y - p[i+1].x*p[i].y;
        t.x += (p[i].x + p[i+1].x) * a;
        t.y += (p[i].y + p[i+1].y) * a;
        A += a;
    }
    t.x /= A*3;
    t.y /= A*3;
    return t;
}
double neiji(point a,point b,point m){
    return (m.x-a.x)*(b.x-a.x)+(m.y-a.y)*(b.y-a.y);
}
int main(){
    int t,i,n;
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        for(i=0;i<n;i++)
            scanf("%lf%lf",&p[i].x,&p[i].y);
        point cen=gravi(p,n);
        tubao(p,n);
        int count=0;
        for(i=0;i<top;i++){
            if(neiji(p[i],p[i+1],cen)>eps&&neiji(p[i+1],p[i],cen)>eps)
                count++;
        }
        printf("%d\n",count);
    }
    return 0;
}