HDU 5733 tetrahedron（计算几何）

题目链接

tetrahedron

题目大意

输入一个四面体求其内心，若不存在内心则输出“O O O O”

解题思路

其实这道题思路很简单，只要类推一下三角形内心公式就可以了。
至于如何判断无解，计算一下体积若V<=0则无解

Code

#include <bits/stdc++.h>

using namespace std;

const double eps = 1e-8;

struct point{
	double x, y, z;
	point (double xx = 0, double yy = 0, double zz = 0){
		x = xx;
		y = yy;
		z = zz;
	}
	friend point operator - (const point &a, const point &b){
		return point(a.x - b.x, a.y - b.y, a.z - b.z);
	}
} p[4];

double cross(point p1, point p2){
	return sqrt(pow(p1.y * p2.z - p1.z * p2.y, 2) + pow(-(p1.x * p2.z - p1.z * p2.x), 2) + pow(p1.x * p2.y - p1.y * p2.x, 2));
}

double vol(point p1, point p2, point p3){
	return fabs((p2.y * p3.z - p2.z * p3.y) * p1.x - (p2.x * p3.z - p2.z * p3.x) * p1.y + (p2.x * p3.y - p2.y * p3.x) * p1.z);
}

int main(){

	while (~scanf("%lf%lf%lf", &p[0].x, &p[0].y, &p[0].z)){
		for (int i = 1; i < 4; i++) scanf("%lf%lf%lf", &p[i].x, &p[i].y, &p[i].z);
	      	
		double V = vol(p[1] - p[0], p[2] - p[0], p[3] - p[0]);
		if (V <= eps){
			printf("O O O O
");
			continue;
		}
		V = V / 6;
		double s1 = 0.5 * cross(p[1] - p[0], p[2] - p[0]);
		double s2 = 0.5 * cross(p[1] - p[0], p[3] - p[0]);
		double s3 = 0.5 * cross(p[2] - p[0], p[3] - p[0]);
		double s4 = 0.5 * cross(p[1] - p[2], p[3] - p[2]);
		double S  = s1 + s2 + s3 + s4;
		double r  = V * 3 / S;
		double x  = (p[0].x * s4 + p[1].x * s3 + p[2].x * s2 + p[3].x * s1) / S;
		double y  = (p[0].y * s4 + p[1].y * s3 + p[2].y * s2 + p[3].y * s1) / S;
		double z  = (p[0].z * s4 + p[1].z * s3 + p[2].z * s2 + p[3].z * s1) / S;
		printf("%.4f %.4f %.4f %.4f
", x, y, z, r);
	}

	return 0;  
}