【POJ - 1269 】Intersecting Lines （计算几何，直线间的位置关系）

题干：

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. 
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. 

Input

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 x1y1x2y2x3y3x4y4. 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).

Output

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".

Sample Input

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

Sample Output

INTERSECTING LINES OUTPUT
POINT 2.00 2.00
NONE
LINE
POINT 2.00 5.00
POINT 1.07 2.20
END OF OUTPUT

题目大意：

   T组样例，每组八个值，前四个值描述一条直线上的两个点，剩下的同理。

解题报告：

  套个模板就可以了。

AC代码：

#include<iostream>
#include<algorithm>
#include<queue>
#include<cstdio>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define ll long long
#define pb push_back
#define pm make_pair
#define fi first
#define se second
using namespace std;
const double eps = 1e-8;
int sgn(double x) {
	if(fabs(x) < eps)return 0;
	if(x < 0) return -1;
	return 1;
}
struct Point {
	double x,y;
	Point(){}
	Point(double x,double y):x(x),y(y){}
	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){}
	pair<Point,int> operator &(const Line &b)const {
        Point res = s;
        if(sgn((s-e)^(b.s-b.e)) == 0)
        {
            if(sgn((b.s-s)^(b.e-s)) == 0)
                return make_pair(res,0);//两直线重合
            else return make_pair(res,1);//两直线平行
        }
        double t = ((s-b.s)^(b.s-b.e))/((s-e)^(b.s-b.e));
        res.x += (e.x - s.x)*t;
        res.y += (e.y - s.y)*t;
        return make_pair(res,2);//有交点
    }
};
double xmult(Point p0,Point p1,Point p2) { //p0p1 X p0p2
	return (p1-p0)^(p2-p0);
}
bool Seg_inter_line(Line l1,Line l2) { //判断直线l1和线段l2是否相交，注意是直线！！如果是线段，还需要判断另一组！！
	return sgn(xmult(l2.s,l1.s,l1.e))*sgn(xmult(l2.e,l1.s,l1.e)) <= 0;
}
double dist(Point a,Point b) {
	return sqrt((b - a)*(b - a));
}
int main()
{
	int t;
	cin>>t;
	double x1,x2,x3,x4,y1,y2,y3,y4;
	puts("INTERSECTING LINES OUTPUT");
	while(t--) {
		scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3,&x4,&y4);
		Line l1 = Line(Point(x1,y1),Point(x2,y2));
		Line l2 = Line(Point(x3,y3),Point(x4,y4));
		pair<Point,int> p = l1&l2; 
		
		if(p.se == 1) puts("NONE");
		else if (p.se == 0) puts("LINE");
		else printf("POINT %.2f %.2f\n",p.fi.x,p.fi.y);
		
	}
	puts("END OF OUTPUT");
	return 0 ;
}