POJ 1269 Intersecting Lines(直线相交判断 求交点)

题目链接

题目大意：每次给你两条线段，判断它们的位置关系，如果相交，还需输出它们的交点。

分析：线段之间存在三种关系，平行，共线，相交。

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<cstdio>
#include<vector>
#include<utility>
#include<cctype>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define inf 0x3f3f3f3f
#define Clear(x) memset(x,0,sizeof(x))
#define fup(i,a,b) for(int i=a;i<b;i++)
#define rfup(i,a,b) for(int i=a;i<=b;i++)
#define fdn(i,a,b) for(int i=a;i>b;i--)
#define rfdn(i,a,b) for(int i=a;i>=b;i--)
typedef long long ll;
using namespace std;
const double pi=acos(-1.0);
const int maxn = 1e2+7;
const double eps = 1e-8;

int sgn(double x)
{
    if(fabs(x)<eps) return 0;
    else return x<0?-1:1;
}

struct Point{
    double x,y;
    Point(){}
    Point(double _x,double _y){x=_x;y=_y;}
};

struct Line{
    Point s,e;
    Line(){}
    Line(Point _s,Point _e){s=_s;e=_e;}
};

double cross(Point a,Point b,Point c)
{
    a.x-=c.x;a.y-=c.y;
    b.x-=c.x;b.y-=c.y;
    return a.x*b.y-a.y*b.x;
}

int parallel(Line l1,Line l2)
{
    return ((l1.e.x-l1.s.x)*(l2.e.y-l2.s.y)-(l1.e.y-l1.s.y)*(l2.e.x-l2.s.x));
}

/*Point intersection(Line l1,Line l2)
{
    Point ret=l1.s;
    double t=((l1.s.x-l2.s.x)*(l2.s.y-l2.e.y)-(l1.s.y-l2.s.y)*(l2.s.x-l2.e.x))/((l1.s.x-l1.e.x)*(l2.s.y-l2.e.y)-(l1.s.y-l1.e.y)*(l2.s.x-l2.e.x));
    ret.x+=(l1.e.x-l1.s.x)*t;
    ret.y+=(l1.e.y-l1.s.y)*t;
    return ret;
}*/

Point intersection(Line l1,Line l2)
{
    Point ret;
    double a1=l1.s.y-l1.e.y,b1=l1.e.x-l1.s.x,c1=l1.s.x*l1.e.y-l1.e.x*l1.s.y;
    double a2=l2.s.y-l2.e.y,b2=l2.e.x-l2.s.x,c2=l2.s.x*l2.e.y-l2.e.x*l2.s.y;
    double x1 = (c1*b2-c2*b1)/(a2*b1-a1*b2);
    double y1 = (a2*c1-a1*c2)/(a1*b2-a2*b1);
    ret.x=x1;ret.y=y1;
    return ret;
}

int main()
{
    int T;
    scanf("%d",&T);
    double x1,y1,x2,y2,x3,y3,x4,y4;
    printf("INTERSECTING LINES OUTPUT\n");
    while(T--)
    {
        Line l1,l2;
        scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&l1.s.x,&l1.s.y,&l1.e.x,&l1.e.y,&l2.s.x,&l2.s.y,&l2.e.x,&l2.e.y);
        if(fabs(parallel(l1,l2))<=eps)
        {
            if(fabs(cross(l1.s,l1.e,l2.s))<=eps)
            {
                printf("LINE\n");
            }
            else printf("NONE\n");
        }
        else{
            Point t = intersection(l1,l2);
            printf("POINT %.2f %.2f\n",t.x,t.y);
        }
    }
    printf("END OF OUTPUT\n");
    return 0;
}