题意：给出N组直线，每组2条直线，求出直线是否相交。如果共线则输出LINE，相交则输入点坐标，否则输出NONE.
分析：模板裸题，直接上模板。。。
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
#include <map>
#include <vector>
#include <set>
#include <string>
#include <math.h> using namespace std; const double eps = 1e-;
const double PI = acos(-1.0);
const int N = ;
int sgn(double x)
{
if(fabs(x) < eps)return ;
if(x < )return -;
else return ;
}
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);
}
//叉积
double operator ^(const Point &b)const
{
return x*b.y - y*b.x;
}
};
struct Line
{
Point s,e;
Line(){}
Line(Point _s,Point _e)
{
s = _s;e = _e;
}
//两直线相交求交点
//第一个值为0表示直线重合，为1表示平行，为0表示相交,为2是相交
//只有第一个值为2时，交点才有意义
pair<int,Point> operator &(const Line &b)const
{
Point res = s;
if(sgn((s-e)^(b.s-b.e)) == )
{
if(sgn((s-b.e)^(b.s-b.e)) == )
return make_pair(,res);//重合
else return make_pair(,res);//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);
}
};
Line seg[];
int main()
{
int T;
scanf("%d",&T);
puts("INTERSECTING LINES OUTPUT");
while(T--)
{
for(int i=;i<=;i++)
{
double a,b,c,d;
scanf("%lf%lf%lf%lf",&a,&b,&c,&d);
seg[i]=Line(Point(a,b),Point(c,d));
}
pair<int,Point> p=seg[]&seg[];
if(p.first==)puts("LINE");
else if(p.first==)puts("NONE");
else
{
printf("POINT %.2lf %.2lf\n",p.second.x,p.second.y);
}
}
puts("END OF OUTPUT");
return ;
}