文章目录
题目描述
平面直角坐标系中有一个点C和一条直线AB,求点C和直线AB的位置关系。
输入描述
输出描述
输入输出样例
输入:
3
0 1
1 0
1 1
0 0
1 1
2 2
0 0
0 1
1 0
输出:
L
IN
R
最终代码c/c++
#include<bits/stdc++.h>
using namespace std;
const double pi = acos(-1.0); //高精度圆周率
const double eps = 1e-8; //偏差值
//判断x是否等于0
int sgn(double x)
{
if(fabs(x) < eps) return 0;
else return x<0?-1:1;
}
struct Point
{
double x, y;
void input(){ scanf("%lf%lf", &x, &y); }
Point(){}
Point(double x,double y):x(x),y(y){}
Point operator + (Point B){return Point(x+B.x,y+B.y);}
Point operator - (Point B){return Point(x-B.x,y-B.y);}
};
//叉积
double Cross(Point A,Point B){return A.x*B.y - A.y*B.x;}
struct Line
{
Point p1,p2; //线上的两个点
Line(){}
Line(Point p1,Point p2):p1(p1),p2(p2){}
};
int Point_line_relation(Point p, Line v)
{
int c = sgn(Cross(p-v.p1,v.p2-v.p1));
if(c < 0)return 1; //1:p在v的左边
if(c > 0)return 2; //2:p在v的右边
return 0; //0:p在v上
}
int main()
{
int t; cin >> t;
while(t--)
{
Point a, b, c;
a.input(); b.input(); c.input();
Line v;
v=Line(a,b);
int pos=Point_line_relation(c, v);
if(pos==1) cout<<"L"<<endl;
if(pos==2)cout<<"R"<<endl;
if(pos==0)cout<<"IN"<<endl;
}
return 0;
}
过程理解
无