1.点绕正轴(x,y,z轴)旋转整理成模板了,代码很短。
2.射线和平面交,定义一个点乘运算符,写起来代码比较短,思路比较清楚。
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
const double eps = 1e-8;
struct Point {
double x, y, z;
Point(double x, double y, double z):
x(x), y(y), z(z){}
Point(){}
Point operator-(const Point &t) const {
return Point(x-t.x, y-t.y, z-t.z);
}
Point operator+(const Point &t) const {
return Point(x + t.x, y + t.y, z + t.z);
}
Point rotz(double t) { // 绕z旋转
return Point(x*cos(t)-y*sin(t), x*sin(t)+y*cos(t), z);
}
Point rotx(double t) { //绕x旋转
return Point(x, y*cos(t)-z*sin(t), y*sin(t)+z*cos(t));
}
Point operator*(const Point &t) const { //叉乘
return Point(y*t.z-z*t.y, z*t.x-x*t.z, x*t.y-y*t.x);
}
Point operator*(const double &t) const {
return Point(x*t, y*t, z*t);
}
double operator^(const Point &t) const { //点乘
return x*t.x+y*t.y+z*t.z;
}
void in() {
scanf("%lf%lf%lf", &x, &y, &z);
}
bool operator<(const Point &t) const {
return y +eps< t.y || (fabs(y-t.y) < eps && x + eps < t.x);
}
void out() {
printf("%lf %lf %lf~~~~\n", x, y, z);
}
}p[103], st[103];
double a, b, c, d;
int n, cnt;
//***********旋转
void rot() {
if(a*a+b*b < eps) return;
double t1 = acos(b/sqrt(a*a+b*b));
double t2 = acos(c/sqrt(a*a+b*b+c*c));
for(int i = 0; i < n; i++)
p[i] = (p[i].rotz(t1)).rotx(t2);
}
//************凸包
double cross(Point a, Point b) {
return a.x*b.y-a.y*b.x;
}
double convex(Point *p, int n) {
sort(p, p+n);
int m = 0;
int i;
for(i = 0; i < n; i++) {
while(m > 1 && cross(st[m-1]-st[m-2],p[i]-st[m-2]) < eps) m--;
st[m++] = p[i];
}
int t = m;
for(i = n-2; i >= 0; i--) {
while(m > t && cross(st[m-1]-st[m-2], p[i]-st[m-2]) < eps) m--;
st[m++] = p[i];
}
double ans = 0;
for(i = 1; i < m-1; i++)
ans += cross(st[i]-st[0], st[i+1]-st[0]);
return fabs(ans)*0.5;
}
int main() {
int i;
while(~scanf("%lf%lf%lf%lf", &a, &b, &c, &d)) {
if(a==0 && b==0 && c==0 && d==0) break;
scanf("%d", &n);
for(i = 0; i <= n; i++) p[i].in();
//*****射线与平面交
Point f(a, b, c);
cnt = 0;
for(i = 0; i < n; i++) {
Point v = p[i]-p[n];
if(fabs(f^v) < eps) continue;
double t = (d-(f^p[n]))/(f^v);
if(t > -eps)
p[cnt++] = v*t+p[n];
}
if(!cnt) puts("0.00");
else if(cnt < n) puts("Infi");
else {
rot();
printf("%.2f\n", convex(p, n));
}
}
return 0;
}