【HDU5733 2016 Multi-University Training Contest 1K】【计算几何 公式做法】tetrahedron 四面体内切球圆心与半径

tetrahedron

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 723    Accepted Submission(s): 294


[align=left]Problem Description[/align]
Given four points ABCD, if ABCD is a tetrahedron, calculate the inscribed sphere of ABCD.
 

[align=left]Input[/align]
Multiple test cases (test cases ≤100).

Each test cases contains a line of 12 integers [−1e6,1e6] indicate
the coordinates of four vertices of ABCD.

Input ends by EOF.

 

[align=left]Output[/align]
Print the coordinate of the center of the sphere and the radius, rounded to 4 decimal places.

If there is no such sphere, output "O O O O".

 

[align=left]Sample Input[/align]

0 0 0 2 0 0 0 0 2 0 2 0
0 0 0 2 0 0 3 0 0 4 0 0

 

[align=left]Sample Output[/align]

0.4226 0.4226 0.4226 0.4226
O O O O

 

[align=left]Author[/align]
HIT
 

[align=left]Source[/align]
2016 Multi-University Training
Contest 1

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre() { freopen("c://test//input.in", "r", stdin); freopen("c://test//output.out", "w", stdout); }
#define MS(x,y) memset(x,y,sizeof(x))
#define MC(x,y) memcpy(x,y,sizeof(x))
#define MP(x,y) make_pair(x,y)
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b>a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b<a)a = b; }
const int N = 0, M = 0, Z = 1e9 + 7, ms63 = 0x3f3f3f3f;
int casenum, casei;
const double eps = 1e-6;
//既可以定义为点，又可以定为为向量
struct Point
{
LL x, y, z;
Point(LL x = 0, LL y = 0, LL z = 0) :x(x), y(y), z(z) {};
bool input()
{
return ~scanf("%lld%lld%lld", &x, &y, &z);
}
Point operator - (Point &b)const
{
return Point(x - b.x, y - b.y, z - b.z);
}
Point operator * (Point &b)const
{
return Point(y * b.z - z * b.y, z * b.x - x * b.z, x * b.y - y * b.x);
}
double length()
{
return sqrt(x*x + y*y + z*z);
}
}p[4];

//已知四面体的6条棱长，求出四面体体积
double volume(double l, double n, double a, double m, double b, double c)
{
double x, y;
x = 4 * a * a * b * b * c * c - a * a * (b * b + c * c - m * m) * (b * b + c * c - m * m) - b * b * (c * c + a * a - n * n) * (c * c + a * a - n * n);
y = c * c * (a * a + b * b - l * l) * (a * a + b * b - l * l) - (a * a + b * b - l * l) * (b * b + c * c - m * m) * (c * c + a * a - n * n);
return sqrt(x - y) / 12;
}

//已知三角形边长，求出三边形面积
double S(double a, double b, double c)
{
double p = (a + b + c) / 2;
return sqrt(p * (p - a) * (p - b) * (p - c));
}

double d[4][4];	//求两点之间的距离
double s[4];	//求该点所对平面的面积
int main()
{
while (p[0].input(), p[1].input(), p[2].input(), p[3].input())
{
for (int i = 0; i < 4; ++i)
{
for (int j = i + 1; j < 4; ++j)d[i][j] = d[j][i] = (p[j] - p[i]).length();
}
double vol = volume(d[0][1], d[0][2], d[0][3], d[1][2], d[1][3], d[2][3]);
double area = 0;
for (int i = 0; i < 4; ++i)
{
vector<double>vt;
for (int j = 0; j < 4; ++j)if (j != i)
{
for (int k = j + 1; k < 4; ++k)if (k != i)vt.push_back(d[j][k]);
}
s[i] = S(vt[0], vt[1], vt[2]);
area += s[i];
}
if (fabs(vol) < eps || fabs(area) < eps)puts("O O O O");
else
{
double x = 0; for (int i = 0; i < 4; ++i)x += s[i] * p[i].x; x /= area;
double y = 0; for (int i = 0; i < 4; ++i)y += s[i] * p[i].y; y /= area;
double z = 0; for (int i = 0; i < 4; ++i)z += s[i] * p[i].z; z /= area;
double r = vol * 3 / area;
printf("%.4f %.4f %.4f %.4f\n", x, y, z, r);
}
}
return 0;
}
/*
【题意】
给你四个点。
让你求四个点是否共面，不共面的话，求出其内切球的球心以及半径。

【类型】
计算几何

【分析】
这题我直接二话不说，
写了一个三分套三分套三分。
但是常数巨大于是TLE了。
还是推公式靠谱。
首先，我们求出该四面体的6条棱长和4个面积之后，
6条棱长可以直接求出四面体体积。
然后(体积*3)/面积之和，就是其内切球的半径。
内切球的圆心公式，有一种非常美的表现形式——
double x = 0; for (int i = 0; i < 4; ++i)x += s[i] * p[i].x; x /= area;
double y = 0; for (int i = 0; i < 4; ++i)y += s[i] * p[i].y; y /= area;
double z = 0; for (int i = 0; i < 4; ++i)z += s[i] * p[i].z; z /= area;
double r = vol * 3 / area;
printf("%.4f %.4f %.4f %.4f\n", x, y, z, r);

即4个点（三维坐标*对面面积）关于总面积的加权平均值。
其实，这个可以是由三角形的内切圆推广而来
内切圆的圆心，
是3个点（三维坐标*对边边长）关于总边长的加权平均值。

【时间复杂度&&优化】
O（1）

【数据】

*/