POJ1269:Intersecting Lines（判断两条直线的关系）

#include <iostream>
#include <string.h>
#include <stdio.h>
#include <algorithm>
#include <math.h>
#include <queue>
#define inf 0x3f3f3f3f
#define eps 1e-9
typedef long long ll;
using namespace std;
int n;
struct Point
{
    double x,y;
};
struct Line
{
    Point a,b;
} line[3];
double xmult(Point p1,Point p2,Point p)//叉积
{
    return (p1.x-p.x)*(p2.y-p.y)-(p1.y-p.y)*(p2.x-p.x);
}
int parallel(Line u,Line v)
{
    return ((u.a.x-u.b.x)*(v.a.y-v.b.y)-(v.a.x-v.b.x)*(u.a.y-u.b.y));
}
Point intersection(Line u,Line v)
{
    Point ret=u.a;
    double t=((u.a.x-v.a.x)*(v.a.y-v.b.y)-(u.a.y-v.a.y)*(v.a.x-v.b.x))/((u.a.x-u.b.x)*(v.a.y-v.b.y)-(u.a.y-u.b.y)*(v.a.x-v.b.x));
    ret.x+=(u.b.x-u.a.x)*t;
    ret.y+=(u.b.y-u.a.y)*t;
    return ret;
}
int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        cout<<"INTERSECTING LINES OUTPUT"<<endl;
        for(int i=0; i<n; i++)
        {
            scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&line[1].a.x,&line[1].a.y,&line[1].b.x,&line[1].b.y,&line[2].a.x,&line[2].a.y,&line[2].b.x,&line[2].b.y);
            if(fabs(parallel(line[1],line[2]))<=eps&&fabs(xmult(line[1].a,line[1].b,line[2].a))<=eps)
            {
                printf("LINE\n");
            }
            else if(fabs(parallel(line[1],line[2]))<=eps)
            {
                printf("NONE\n");
            }
            else
            {
                Point tt=intersection(line[1],line[2]);
                printf("POINT %.2f %.2f\n", tt.x,tt.y);//poj只能输出f
            }
        }
        printf("END OF OUTPUT\n");
    }
    return 0;
}