UVA - 378 Intersecting Lines【基础计算几何】

【题目描述】
We all know that a pair of distinct points on a plane defines a line and that a pair of lines on a plane
will intersect in one of three ways: 1) no intersection because they are parallel, 2) intersect in a line
because they are on top of one another (i.e. they are the same line), 3) intersect in a point. In this
problem you will use your algebraic knowledge to create a program that determines how and where two lines intersect.
我们都知道，平面上的一对不同的点定义了一条线，平面上的一对线将以三种方式之一相交：(1)没有交叉，因为它们是平行的，2)在一条线上相交，因为它们是相交的（他们是同一条直线）。(3)点相交。在这个问题中，您将使用您的代数知识来创建一个程序，该程序确定两行交叉的方式和位置。
Your program will repeatedly read in four points that define two lines in the x-y plane and determine how and where the lines intersect. All numbers required by this problem will be reasonable, say between -1000 and 1000.
您的程序将重复读取四个点，这些点定义了x-y平面上的两行，并确定了这些线是如何和在何处相交的。这个问题所要求的所有数字都是合理的，比如说-1000到1000之间。

【输入】
The first line contains an integer N between 1 and 10 describing how many pairs of lines are represented.The next N lines will each contain eight integers. These integers represent the coordinates of four points on the plane in the order x1 y1 x2 y2 x3 y3 x4 y4. Thus each of these input lines represents two lines on the plane: the line through (x1, y1) and (x2, y2) and the line through (x3, y3) and (x4, y4). The point (x1, y1) is always distinct from (x2, y2). Likewise with (x3, y3) and (x4, y4).
第一行包含介于1到10之间的整数N，描述代表了多少对行。接下来的N行将包含8个整数。这些整数以x1y1x2y2x3y3x4y4的顺序表示平面上四个点的坐标。因此，这些输入线中的每一行代表平面上的两条线：穿过(x1，y1)和(x2，y2)的线和穿过(x3，y3)和(x4，y4)的线。点(x1，y1)总是与(x2，y2)不同。(x3，y3)和(x4，y4)也是如此。

【输出】
There should be N + 2 lines of output. The first line of output should read ‘INTERSECTING LINES
OUTPUT’. There will then be one line of output for each pair of planar lines represented by a line of
input, describing how the lines intersect: none, line, or point. If the intersection is a point then your
program should output the x and y coordinates of the point, correct to two decimal places. The final
line of output should read ‘END OF OUTPUT’.
有N+2行输出。第一行输出‘INTERSECTING LINES OUTPUT’。然后，每一对由输入线表示的平面线将有一行输出，描述这些线是如何相交的：无、线或点。如果交点是一个点，那么程序应该输出点的x和y坐标，精确到小数点后两位。最后一行输出‘END OF OUTPUT’。

【样例输入】
5
0 0 4 4 0 4 4 0
5 0 7 6 1 0 2 3
5 0 7 6 3 -6 4 -3
2 0 2 27 1 5 18 5
0 3 4 0 1 2 2 5

【样例输出】
INTERSECTING LINES OUTPUT
POINT 2.00 2.00
NONE
LINE
POINT 2.00 5.00
POINT 1.07 2.20
END OF OUTPUT

题目链接：https://cn.vjudge.net/problem/UVA-378

基础计算几何，判断两直线关系

代码如下：

#include <iostream>
#include <iomanip>
#include <cmath>
using namespace std;
static const double EPS=1e-8;
int sgn(double x)
{
    if(fabs(x)<EPS) return 0;
    if(x<0) return -1;
    else return 1;
}
struct Point{
    double x,y;
    Point(){}
    Point(double _x,double _y):x(_x),y(_y){}
    void input() { cin>>x>>y; }
    void output() { cout<<fixed<<setprecision(2)<<x<<" "<<y<<endl; }
    Point operator - (const Point &b) const { return Point(x-b.x,y-b.y); }
    double operator ^ (const Point &b) const { return x*b.y-y*b.x; }
    double operator * (const Point &b) const { return x*b.x+y*b.y; }
};
struct Line{
    Point s,e;
    Line(){}
    Line(Point _s,Point _e):s(_s),e(_e){}
    void input() { s.input(); e.input(); }
    bool parallel(Line v) { return sgn((e-s)^(v.e-v.s))==0; }
    int relation(Point p) {
        int c=sgn((p-s)^(e-s));
        if(c<0) return 1;
        else if(c>0) return 2;
        else return 3;
    }
    int linecrossline(Line v) {
        if((*this).parallel(v))
            return v.relation(s)==3;
        return 2;
    }
    Point crosspoint(Line v) {
        double a1=(v.e-v.s)^(s-v.s),a2=(v.e-v.s)^(e-v.s);
        return Point((s.x*a2-e.x*a1)/(a2-a1),(s.y*a2-e.y*a1)/(a2-a1));
    }
};
int main()
{
    std::ios::sync_with_stdio(false);
    std::cin.tie(0),cout.tie(0);
    int T;
    cin>>T;
    cout<<"INTERSECTING LINES OUTPUT"<<endl;
    while(T--)
    {
        Line line1,line2;
        line1.input(),line2.input();
        if(line1.linecrossline(line2)==0)
            cout<<"NONE"<<endl;
        else if(line1.linecrossline(line2)==1)
            cout<<"LINE"<<endl;
        else
        {
            cout<<"POINT ";
            Point p=line1.crosspoint(line2);
            p.output();
        }
    }
    cout<<"END OF OUTPUT"<<endl;
    return 0;
}