HDU5733 tetrahedron

计算几何题给出四个点求这四个点构成三棱锥的内切球的三维坐标和半径

先判断这个四个点是否共面共面就直接输出O O O O 然后就用公式啦

注意scanf的时候要用%lf 最开始用的%f就各种出现nan和inf

这里贴个链接公式的链接 http://www.docin.com/p-504197705.html?qq-pf-to=pcqq.c2c

AC代码：

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 typedef unsigned long long ull;
 5 
 6 class point{
 7 public:
 8     double x;
 9     double y;
10     double z;
11     point(double xx = 0, double yy = 0, double zz = 0){
12         x = xx;
13         y = yy;
14         z = zz;
15     }
16 };
17 
18 point operator -(point a, point b){
19     return point(a.x - b.x, a.y - b.y, a.z - b.z);
20 }
21 
22 point cross_product(point a, point b){
23     return point(a.y * b.z - a.z * b.y, a.z * b.x - a.x * b.z, a.x * b.y - a.y * b.x);
24 }
25 
26 double dot_product(point a, point b){
27     return a.x * b.x + a.y * b.y + a.z * b.z;
28 }
29 
30 double Herons_formula(double a, double b, double c){
31     double p = (a + b + c) / 2;
32     return sqrt(p * (p - a) * (p - b) * (p - c));
33 }
34 
35 double dist(point a, point b){
36     return 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));
37 }
38 
39 int main(){
40     point A, B, C, D;
41     point AB, AC, AD;
42     double s1, s2, s3, s4;
43     double ab, ac, ad, bc, bd, cd;
44     double x, y, z, r;
45     double V;
46     while (scanf("%lf%lf%lf%lf%lf%lf%lf%lf%lf%lf%lf%lf", &A.x, &A.y, &A.z, &B.x, &B.y, &B.z, &C.x, &C.y, &C.z, &D.x, &D.y, &D.z) != EOF){
47         AB = A - B;
48         AC = A - C;
49         AD = A - D;
50         V = abs(dot_product(cross_product(AB , AC), AD) / 6);
51         if (V == 0){
52             cout<<"O O O O"<<endl;
53             continue;
54         }
55         ab = dist(A, B);
56         ac = dist(A, C);
57         ad = dist(A, D);
58         bc = dist(B, C);
59         bd = dist(B, D);
60         cd = dist(C, D);
61         s1 = Herons_formula(bc, bd, cd);
62         s2 = Herons_formula(ac, ad, cd);
63         s3 = Herons_formula(ab, ad, bd);
64         s4 = Herons_formula(ab, ac, bc);
65         x = (s1 * A.x + s2 * B.x + s3 * C.x + s4 * D.x) / (s1 + s2 + s3 + s4);
66         y = (s1 * A.y + s2 * B.y + s3 * C.y + s4 * D.y) / (s1 + s2 + s3 + s4);
67         z = (s1 * A.z + s2 * B.z + s3 * C.z + s4 * D.z) / (s1 + s2 + s3 + s4);
68         r = (V * 3) / (s1 + s2 + s3 + s4);
69         printf("%.4lf %.4lf %.4lf %.4lf\n", x, y, z, r);
70     }
71     return 0;
72 }

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