【计算几何入门】poj 1269

嗯终于决定开始补计算几何了

从最基础的题目开始做

这题意思就是判断线段的三种位置关系没什么可说的

因为输出double用了 lf 调试了很长时间。。

哦对有个问题，那就是这个题目好像没有考虑两个线段首尾相连且斜率相同的情况？

本题的收获：计算几何因为对精度的要求很高，所以不能用解析几何的办法求斜率什么的 ，必须化除法为乘法，这样才能保证精度。

其次，能用叉乘的地方尽量用叉积，真的是非常好用的运算！

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#define rep(i, j, k) for(int i = j; i <= k; i++)
#define vector point
#define eps 1e-8

using namespace std;

struct point
{
double x, y;
point (double x = 0, double y = 0) : x (x), y (y) {}
};
struct line
{
point a, b;
};

vector operator + (point a, point b)
{
return vector (a.x + b.x, a.y + b.y);
}
vector operator - (point a, point b)
{
return vector (a.x - b.x, a.y - b.y);
}
vector operator * (point a, double b)
{
return vector (a.x * b, a.y * b);
}
vector operator / (point a, double b)
{
return vector (a.x / b, a.y / b);
}
bool operator == (point a, point b)
{
return fabs (a.x - b.x) < eps && fabs (a.y - b.y) < eps;
}

double length (vector x)
{
return sqrt (x.x * x.x + x.y * x.y);
}

double cross (vector a, vector b)
{
//	printf ("cross %.2lf\n", a.x * b.y - a.y * b.x);
return a.x * b.y - a.y * b.x;
}

point GetLineIntersection (point p, vector v, point q, vector w)
{
vector u = p - q;
double t = cross (w, u) / cross (v, w);
//print (p), print (v); printf ("length ------------%.2lf\n", t); print (p + v * t);
return p + v * t;
}

void print (point x)
{
printf ("%.2lf %.2lf\n", x.x, x.y);
}

int main()
{
int Time;
cout << "INTERSECTING LINES OUTPUT" << endl;
cin >> Time;
while (Time--)
{
line u, v;
scanf ("%lf%lf%lf%lf%lf%lf%lf%lf", &u.a.x ,&u.a.y, &u.b.x, &u.b.y, &v.a.x, &v.a.y, &v.b.x, &v.b.y);
if (fabs (cross (u.b - u.a, v.b - v.a)) < eps)
{
if (fabs (cross (u.b - u.a, v.a - u.a)) < eps)
printf ("LINE\n");
else
printf ("NONE\n");
continue;
}
point p = u.a, q = v.a;
vector vv = u.b - u.a, w = v.b - v.a;
vv = vv / length (vv);
w = w / length (w);
point ans = GetLineIntersection (p, vv, q, w);
printf ("POINT %.2f %.2f\n", ans.x, ans.y);
}
printf ("END OF OUTPUT\n");
return 0;
}