hdu 5733 tetrahedron 四面体内切球球心公式

tetrahedron

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 889    Accepted Submission(s): 382

Problem Description

Given four points ABCD, if ABCD is a tetrahedron, calculate the inscribed sphere of ABCD.

 

Input

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.

 

Output

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".

 

Sample Input

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

 

Sample Output

0.4226 0.4226 0.4226 0.4226
O O O O

 

Author

HIT

 

Source

2016 Multi-University Training Contest 1

 

题意：给你四个点的坐标，判断是否有内切圆，如果有内切圆的话，输出内切圆圆心的坐标和半径；

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <map>
#include <vector>
#include <queue>
#include <cstring>
#include <string>
#include <algorithm>
using namespace std;
typedef  long long  ll;
typedef unsigned long long ull;
#define MM(a,b) memset(a,b,sizeof(a));
#define inf 0x7f7f7f7f
#define FOR(i,n) for(int i=1;i<=n;i++)
#define CT continue;
#define PF printf
#define SC scanf
const int mod=1000000007;
const int N=1e6+100;
int n,m,c[N],pre[N],sum[N],a[N],ans[N];

struct Point{
  ll x,y,z;
  void read()
  {
      scanf("%lld%lld%lld",&x,&y,&z);
  }
}p[6];

Point operator-(Point a,Point b)
{
   return (Point){b.x-a.x,b.y-a.y,b.z-a.z};
}

double dis(Point a)
{
   return sqrt(a.x*a.x+a.y*a.y+a.z*a.z);
}

Point cross(Point a,Point b)
{
    return (Point){a.y*b.z-b.y*a.z,a.z*b.x-a.x*b.z,a.x*b.y-a.y*b.x};
}

double dot(Point a,Point b)
{
    return a.x*b.x+a.y*b.y+a.z*b.z;
}

double pointtoface(Point c,Point a,Point b,Point d)
{
    Point m=cross(b-a,d-a);
    return dot(m,c-a)/dis(m);
}//三维几何中点到面的距离利用 向量a*向量b=|a|*|b|cos(ang)

int main()
{
    double s[5];
    while(~scanf("%lld%lld%lld",&p[1].x,&p[1].y,&p[1].z))
    {
        p[2].read();p[3].read();p[4].read();
        if(dot(cross(p[2]-p[1],p[3]-p[1]),p[4])==0) {printf("O O O O\n");CT;}
        double ts=0;
        s[1]=dis(cross(p[3]-p[2],p[4]-p[2]))/2;
        s[2]=dis(cross(p[3]-p[1],p[4]-p[1]))/2;
        s[3]=dis(cross(p[2]-p[1],p[4]-p[1]))/2;
        s[4]=dis(cross(p[3]-p[1],p[2]-p[1]))/2;
        for(int i=1;i<=4;i++) ts+=s[i];

        double h=pointtoface(p[3],p[1],p[2],p[4]);
        double r=fabs(s[3]/ts*h);

        double x=(s[1]*p[1].x+s[2]*p[2].x+s[3]*p[3].x+s[4]*p[4].x)/ts;
        double y=(s[1]*p[1].y+s[2]*p[2].y+s[3]*p[3].y+s[4]*p[4].y)/ts;
        double z=(s[1]*p[1].z+s[2]*p[2].z+s[3]*p[3].z+s[4]*p[4].z)/ts;

        printf("%.4f %.4f %.4f %.4f\n",x,y,z,r);
    }
    return 0;
}

　　内切球坐标公式：http://www.docin.com/p-504197705.html?qq-pf-to=pcqq.c2c

这道题目了解下内切球球心公式就好，其他没有什么难的