Poj Intersecting Lines （求直线交点）

• 思路
  有两个公式
  1.已知两点坐标(x1,y1,)，(x2,y2).求直线方程(一般形式)
  $Ax+By+C=0$
  $A=y2-y1,B=x1-x2,C=x2\ast y1-x1\ast y2$
  2.已知两直线方程(一般形式)
  $A_{1}x+B_{1}y+C_1=0$
  $A_{2}x+B_{2}y+C_2=0$
  求两个直线的交点( 有交点的情况下)
  $x=\frac{B_1\ast C_2-B_2\ast C_1}{A_1\ast B_2-A_2\ast B_1}$
  $y=\frac{A_2\ast C_1-A_1\ast C_2}{A_1\ast B_2-A_2\ast B_1}$
• 代码

#pragma GCC optimize(2)
//#include<bits/stdc++.h>
#include<iostream>
#include<cstdio>
#include<cmath> 

using namespace std;

typedef long long ll;
typedef unsigned long ul;
typedef unsigned long long ull;
#define pi acos(-1.0)
#define e exp(1.0)
#define pb push_back
#define mk make_pair
#define fir first
#define sec second
#define scf scanf
#define prf printf
typedef pair<ll,ll> pa;
const ll INF=0x3f3f3f3f3f3f3f3f;
int N,T;
const double eps=1e-8;
double x1,y1,x2,y2,x3,y3,x4,y4;
void do_(){
    
    
	int i,j,x_1,y_1,x_2,y_2;
	x_1=x2-x1;
	y_1=y2-y1;
	x_2=x4-x3;
	y_2=y4-y3;
	if(fabs(x_1*y_2-x_2*y_1)<eps){
    
    
		x_1=x3-x1;
		y_1=y3-y1;
		x_2=x4-x1;
		y_2=y4-y1;
		if(fabs(x_1*y_2-x_2*y_1)<eps)
		prf("LINE\n");
		else
		prf("NONE\n");
	}
	else{
    
    
		double a1=y2-y1,b1=x1-x2,c1=x2*y1-x1*y2;
		double a2=y4-y3,b2=x3-x4,c2=x4*y3-x3*y4;
		double x=(b1*c2-b2*c1)/(a1*b2-a2*b1);
		double y=(a2*c1-a1*c2)/(a1*b2-a2*b1);
		prf("POINT %.2f %.2f\n",fabs(x)<eps?0:x,fabs(y)<eps?0:y);
	}
	return ; 
}
int main()
{
    
    
//  freopen(".../.txt","w",stdout);
//  freopen(".../.txt","r",stdin);
//	ios::sync_with_stdio(false);
//	cin>>N;
	scf("%d",&N);
	prf("INTERSECTING LINES OUTPUT\n");
	while(N--){
    
    
		scf("%lf %lf %lf %lf %lf %lf %lf %lf",&x1,&y1,&x2,&y2,&x3,&y3,&x4,&y4);
		do_();
	}
	prf("END OF OUTPUT\n");
	return 0;
}