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;
}