51NOD1265(四点共面)

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

=(

)，

=(

),a×b=(

-

)i+(

-

)j+(

-

)k,

写成行列式形式 .点积公式为

和

 , x1 * x2 + y1 * y2,自行扩展至三维.

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 DoubleQ    freopen("in.txt","r",stdin);#endif    int 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");            else                printf("No\n");    }    return 0;}/***************************************************   Copyright By DoubleQ*   Written in 2015*   Blog Address : zhanghe.ac.cn*                  http://blog.csdn.net/u013447865*   Email Address: [e-mail]**************************************************/