HDOJ1411 校慶神祕建築/2015HDU Summer Trainning(3) - Team 1008 （幾何）

HDOJ1411：按照一定順序給出四面體的六條邊長，求四面體體積。

公式題目：

計算的時候注意把邊和角對應好，sqrt裏面注意浮點誤差，不要對負數開方。

#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
using namespace std;
double a,b,c,d,e,f;

double CalAng(double a,double b,double c)
{
    double ans=(a*a+b*b-c*c)/(2.0*a*b);
    ans=acos(ans);
    return ans;
}

void work()
{
    double W,A,B,C,V;
    A=CalAng(a,b,d);
    B=CalAng(b,c,f);
    C=CalAng(a,c,e);
    W=(A+B+C)/2.0;
    V=a*b*c*sqrt(sin(W)*sin(W-A)*sin(W-B)*sin(W-C));
    V/=3.0;
    printf("%0.4lf\n",V);

}

int main()
{
    while(scanf("%lf%lf%lf%lf%lf%lf",&a,&b,&c,&d,&e,&f)==6)
        work();
    return 0;
}

同理，組隊賽第三場那道題：

已知一個三角形ABC三點在空間內的座標，和空間某一點q到ABC三點的距離，求q點到ABC的距離。

同樣方法求出體積再求底面積高就出來了。

#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
using namespace std;

struct Point
{
	double x;
	double y;
	double z;
};

double dist(Point a, Point b)
{
	double res = sqrt((a.x - b.x)*(a.x - b.x) + (a.y - b.y)*(a.y - b.y) + (a.z - b.z)*(a.z - b.z));
	return res;
}

double CalAng(double a, double b, double c)
{
	double ans = (a*a + b*b - c*c) / (2.0*a*b);
	ans = acos(ans);
	return ans;
}

double eps_sin(double x)
{
	double res = sin(x);
	if (fabs(res) < 1e-6) res = 0;
	return res;
}

double CalArea(double a, double b, double c)
{
	double p = (a + b + c) / 2.0;
	double res = sqrt(fabs(p*(p-a)*(p-b)*(p-c)));
	return res;
}

void work()
{
	Point P[4];
	double a, b, c, d, e, f,h;
	double W, A, B, C, V;

	for (int i = 1; i<4; i++)
		scanf("%lf%lf%lf", &P[i].x, &P[i].y, &P[i].z);
	scanf("%lf%lf%lf",&a, &b, &c);

	d = dist(P[1], P[2]);
	e = dist(P[1], P[3]);
	f = dist(P[2], P[3]);

	A = CalAng(a, b, d);
	B = CalAng(b, c, f);
	C = CalAng(a, c, e);
	W = (A + B + C) / 2.0;
	V = a*b*c*sqrt(eps_sin(W)*eps_sin(W - A)*eps_sin(W - B)*eps_sin(W - C));
	h = V / CalArea(d,e,f);
	printf("%0.2lf\n",h);

}

int main()
{
	int T;
	scanf("%d", &T);
	while (T--)  work();
	return 0;
}