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题目链接:点击打开链接
解题思路:
判断四点共面,先求出三点构成的平面的法向量(叉积),如果第四个点和前三点任意一点构成的向量与平面法向量垂直(点积为0),则四点共面.
回忆下叉积和点积.对于三位空间向量,叉积公式为
写成行列式形式
.点积公式为
完整代码:
#include <algorithm>
#include <iostream>
#include <cstring>
#include <climits>
#include <cstdio>
#include <string>
#include <cmath>
#include <set>
#include <queue>
#include <map>
#include <vector>
#include <cstdlib>
#include <stack>
#include <time.h>
using namespace std;
typedef long long LL;
const int MOD = int(1e9)+7;
const int INF = 0x3f3f3f3f;
const double EPS = 1e-9;
const double PI = acos(-1.0); //M_PI;
const int maxn = 100001;class Point_3
{
public:double x , y , z;Point_3() {}Point_3(double xx , double yy , double zz) : x(xx) , y(yy) , z(zz) {}void input(){scanf("%lf%lf%lf",&x,&y,&z);}friend Point_3 operator - (const Point_3 &a , const Point_3 &b){return Point_3(a.x - b.x , a.y - b.y , a.z - b.z);}
};Point_3 det(const Point_3 &a , const Point_3 &b)
{return Point_3(a.y * b.z - a.z * b.y , a.z *b.x - a.x * b.z , a.x * b.y - a.y * b.x);
}double dot(const Point_3 &a , const Point_3 &b)
{return a.x * b.x + a.y * b.y + a.z * b.z;
}Point_3 pvec(Point_3 &s1 , Point_3 &s2 , Point_3 s3)
{return det((s1 - s2) , (s2 - s3));
}bool zreo(double x)
{return fabs(x) < EPS;
}int dots_onplane(Point_3 a , Point_3 b , Point_3 c , Point_3 d )
{return zreo(dot(pvec(a , b , c ) , d - a));
}int main()
{
#ifdef DoubleQfreopen("in.txt","r",stdin);
#endifint T;scanf("%d",&T);while(T--){Point_3 a , b , c , d;a.input();b.input();c.input();d.input();if(dots_onplane(a , b , c , d))printf("Yes\n");elseprintf("No\n");}return 0;
}/*************************************************
*
* Copyright By DoubleQ
* Written in 2015
* Blog Address : zhanghe.ac.cn
* http://blog.csdn.net/u013447865
* Email Address: acmer_doubleq@qq.com
*
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