POJ 1269 (直线求交）

Problem Intersecting Lines (POJ 1269)

题目大意

　　给定两条直线，问两条直线是否重合，是否平行，或求出交点。

解题分析

　　主要用叉积做，可以避免斜率被0除的情况。

　　求交点P0: 已知P1 P2 P3 P4

　　运用 P0P1 X P0P2 = 0 和 P0P3 X P0P4 = 0 

　　C++ 用%.2lf g++ 用 %.2f!!!

　　C++ 用%.2lf g++ 用 %.2f!!!

　　C++ 用%.2lf g++ 用 %.2f!!!

参考程序

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;

#define eps 1e-8

struct P{
    double x,y;
    
    P(){}
    P(double x,double y):x(x),y(y){}

    friend P operator +(P a,P b){
        return P(a.x+b.x,a.y+b.y);
    } 
    friend P operator -(P a,P b){
        return P(a.x-b.x,a.y-b.y);
    } 
    friend double operator *(P a,P b){
        return a.x * b.y - a.y * b.x ;
    }
    friend double operator /(P a,P b){
        return a.x * b.x + a.y * b.y ;
    }
    friend bool operator ==(P a,P b){
        return fabs(a.x-b.x)<eps && fabs(a.y-b.y)<eps;
    }
    friend bool operator !=(P a,P b){
        return !(a==b);
    }
    friend bool operator <(P a,P b){
        if (fabs(a.y-b.y)<eps) return a.x<b.x;
        return a.y<b.y;
    }
    friend double turn(P a,P b,P c){    //向量AB X 向量AC
        return (b-a)*(c-a);
    }
    friend void print(P a){
        printf("%.2lf %.2lf
",a.x,a.y);
    }
};

struct L{
    P a,b;
    L(){}
    L(P a,P b):a(a),b(b){}
    friend P subl(L a){  //向量l
        return P(a.b.x-a.a.x,a.b.y-a.a.y);
    }
};

bool online(L l,P p){
    if (fabs(turn(l.a,l.b,p))<eps) return 1;
    return 0;
}
P solve(double a1,double b1,double c1,double a2,double b2,double c2){
    P a;
    a.x = (b1*c2-b2*c1)/(a1*b2-a2*b1);
    a.y = (a1*c2-a2*c1)/(a2*b1-a1*b2);
    return a;
}
void check(L lx,L ly){
    if (online(lx,ly.a) && online(lx,ly.b)) printf("LINE
"); else
    if (fabs(subl(lx)*subl(ly))<eps) printf("NONE
");     else
    {
        double x1=lx.a.x , y1=lx.a.y;
        double x2=lx.b.x , y2=lx.b.y;
        double x3=ly.a.x , y3=ly.a.y;
        double x4=ly.b.x , y4=ly.b.y;
        double a1 = y1 - y2 , b1 = x2 - x1 , c1 = x1*y2 - x2*y1 ;
        double a2 = y3 - y4 , b2 = x4 - x3 , c2 = x3*y4 - x4*y3 ;
        printf("POINT ");
        print(solve(a1,b1,c1,a2,b2,c2));
    }
}

int main(){
    int T;
    scanf("%d",&T);
    printf("INTERSECTING LINES OUTPUT
");
    while (T--){
        double x1,y1,x2,y2,x3,y3,x4,y4;
        scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3,&x4,&y4);
        P a(x1,y1),b(x2,y2),c(x3,y3),d(x4,y4);
        L lx(a,b),ly(c,d);
        check(lx,ly);
    }
    printf("END OF OUTPUT
");
}