四面体内接圆圆心的坐标计算模板
设四面体的四个点为
所对的面的面积为
顶点的坐标为
那么四面体的內接圆圆心的坐标:
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<cmath>
#include<vector>
#include<set>
#include<queue>
#include<limits.h>
#include<string.h>
#include<map>
#include<list>
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long long LL;
#define inf int(0x3f3f3f3f)
#define mod ll(1e9+7)
#define eps double(1e-7)
#define pi acos(-1.0)
#define lson root << 1
#define rson root << 1 | 1
struct Point
{
ll x,y,z;
} mi[20];
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 {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;
}
int main()
{
while(scanf("%lld%lld%lld",&mi[1].x,&mi[1].y,&mi[1].z)!=EOF)
{
for(int i=2; i<=4; i++)
cin>>mi[i].x>>mi[i].y>>mi[i].z;
if(dot(cross(mi[2]-mi[1],mi[3]-mi[1]),mi[4])==0)
{
printf("No such condition!");
continue;
}
double s1=dis(cross(mi[3]-mi[2],mi[4]-mi[2]))/2.0;
double s2=dis(cross(mi[3]-mi[1],mi[4]-mi[1]))/2.0;
double s3=dis(cross(mi[2]-mi[1],mi[4]-mi[1]))/2.0;
double s4=dis(cross(mi[3]-mi[1],mi[2]-mi[1]))/2.0;
double rx=(s1*mi[1].x+s2*mi[2].x+s3*mi[3].x+s4*mi[4].x)/(s1+s2+s3+s4);
double ry=(s1*mi[1].y+s2*mi[2].y+s3*mi[3].y+s4*mi[4].y)/(s1+s2+s3+s4);
double rz=(s1*mi[1].z+s2*mi[2].z+s3*mi[3].z+s4*mi[4].z)/(s1+s2+s3+s4);
printf("%.4f %.4f %.4f\n",rx,ry,rz);
}
}