Ampzz 2011 Cross Spider 计算几何

原题链接：http://codeforces.com/gym/100523/attachments/download/2798/20142015-ct-s02e07-codeforces-trainings-season-2-episode-7-en.pdf

题意

给你一堆三维点，问你他们是否共面

题解

模板题，套版就好，需要注意的是共线

代码

#include<iostream>
#include<cmath>
#include<cstdio>
#define MAX_N 100005
using namespace std;
 
const double eps=1e-;
 
inline double Sqrt(double a) {
    return a <=  ?  : sqrt(a);
}
 
inline double Sqr(double a) {
    return a * a;
}
 
class Point_3 {
public:
    double x, y, z;
 
    Point_3() { }
 
    Point_3(double xx, double yy, double zz) : x(xx), y(yy), z(zz) { }
 
    Point_3 operator-(Point_3 a) {
        return Point_3(x - a.x, y - a.y, z - a.z);
    }
 
    double Length() const {
        return Sqrt(Sqr(x) + Sqr(y) + Sqr(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;
}
 
double vlen(Point_3 P){return P.Length();}
 
int dot_inline(Point_3 p1,Point_3 p2,Point_3 p3) {
    return vlen(Det(p1 - p2, p2 - p3)) < eps;
}
 
bool zero(double x) {
    return fabs(x) < eps;
}
 
Point_3 pvec(Point_3 s1,Point_3 s2,Point_3 s3) {
    return Det((s1 - s2), (s2 - s3));
}
 
int dots_onplane(Point_3 a,Point_3 b,Point_3 c,Point_3 d) {
    return zero(Dot(pvec(a, b, c), d - a));
}
 
int n;
Point_3 point3[MAX_N];
 
int main() {
    scanf("%d", &n);
    for (int i = ; i < n; i++) {
        int a, b, c;
        scanf("%d%d%d", &a, &b, &c);
        point3[i] = Point_3(a, b, c);
    }
    if (n <= ) {
        cout << "TAK" << endl;
        return ;
    }
    int p = -;
    for (int i = ; i < n; i++) {
        if (dot_inline(point3[], point3[], point3[i]) == ) {
            p = i;
            break;
        }
    }
    if (p == -) {
        cout << "TAK" << endl;
        return ;
    }
    for (int i = ; i < n; i++) {
        if (i == p)continue;
        if (dots_onplane(point3[], point3[], point3[p], point3[i]) == ) {
            cout << "NIE" << endl;
            return ;
        }
    }
    cout << "TAK" << endl;
    return ;
}