题目链接 tetrahedron 题目大意 输入一个四面体求其内心,若不存在内心则输出"O O O O" 解题思路 其实这道题思路很简单,只要类推一下三角形内心公式就可以了. 至于如何判断无解,计算一下体积若V<=0则无解 Code #include <bits/stdc++.h> using namespace std; const double eps = 1e-8; struct point{ double x, y, z; point (double xx =…

tetrahedron 传送门 Time Limit: 2000/1000 MS (Java/Others)   Memory Limit: 65536/65536 K (Java/Others) Problem DescriptionGiven four points ABCD, if ABCD is a tetrahedron, calculate the inscribed sphere of ABCD. InputMultiple test cases .Each test cases…

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 spher…

HDU 4998 Rotate (计算几何) 题目链接http://acm.hdu.edu.cn/showproblem.php?pid=4998 Description Noting is more interesting than rotation! Your little sister likes to rotate things. To put it easier to analyze, your sister makes n rotations. In the i-th time, s…

/* hdu 4643 GSM 计算几何 - 点线关系 N个城市,任意两个城市之间都有沿他们之间直线的铁路 M个基站 问从城市A到城市B需要切换几次基站 当从基站a切换到基站b时,切换的地点就是ab的中垂线与铁路的交点(记录由哪两个基站得到的交点,方便切换)处 枚举任意两个基站与铁路的交点,按到城市A的距离排序 求出在城市A时用的基站j,然后开始遍历交点,看从j可以切换到哪个基站(假设是k),然后再看可以从k可以切换到哪个基站 */ #include<stdio.h> #include<…

Special Tetrahedron 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5839 Description Given n points which are in three-dimensional space(without repetition). Please find out how many distinct Special Tetrahedron among them. A tetrahedron is called Sp…

输入4个点三维坐标,如果是六面体,则输出内切球的球心坐标和半径. 点pi对面的面积为si,点a,b,c组成的面积=|ab叉乘ac|/2. 内心为a,公式: s0=s1+s2+s3+s4 a.x=∑si*pi.x/s0 a.y=∑si*pi.y/s0 a.z=∑si*pi.z/s0 n为p1.p2.p3的法向量,n=p1p2叉乘p1p3 半径=p1a点乘n/|n| #include <cstdio> #include <cmath> #define dd double struct…

tetrahedron/center> 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5726 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 lin…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1174 解题报告:就是用到了三维向量的点积来求点到直线的距离,向量(x1,y1,z1)与(x2,y2,z2)的点积是:x1*x2+y1*y2+z1*z2. 然后要注意的就是当两个向量的夹角大于等于90度时,无论如何都不能射中. #include<cstdio> #include<cstring> #include<iostream> #include<algorithm…

链接:http://acm.hdu.edu.cn/showproblem.php?pid=1086 You can Solve a Geometry Problem too Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 7167    Accepted Submission(s): 3480 Problem Description Ma…