Parallel/Orthogonal (线段几何1)

最新推荐文章于 2021-11-06 22:45:16 发布

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本文介绍了一种算法，用于判断二维平面上两直线是否平行或正交。通过输入四点坐标，算法能够准确判断两条直线的关系，并输出相应结果。使用了斜率比较和复数乘法两种方法实现。

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Parallel/Orthogonal

For given two lines s1 and s2, print "2" if they are parallel, "1" if they are orthogonal, or "0" otherwise.

s1 crosses points p0 and p1, and s2 crosses points p2 and p3.

Input

The entire input looks like:

q (the number of queries)
1st query
2nd query
...
qth query

Each query consists of integer coordinates of the points p0, p1, p2, p3 in the following format:

xp0 yp0 xp1 yp1 xp2 yp2 xp3 yp3

Output

For each query, print "2", "1" or "0".

Constraints

1 ≤ q ≤ 1000
-10000 ≤ xpi, ypi ≤ 10000
p0≠p1 and p2≠p3.

Sample Input 1

3
0 0 3 0 0 2 3 2
0 0 3 0 1 1 1 4
0 0 3 0 1 1 2 2

Sample Output 1

2
1
0

题目连接：http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=CGL_2_A

#include<string.h>
#include<algorithm>
#include<stdio.h>
using namespace std;
int t;
double x1,y1,x2,y2,x3,y3,x4,y4;
int main()
{
    for(scanf("%d",&t);t;--t)
    {
        scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3,&x4,&y4);
        double k1=(x1-x2)/(y1-y2);
        double k2=(x3-x4)/(y3-y4);
        double b1=y1-k1*x1;
        double b2=y3-k2*x3;
        if((k1-k2<0.0000001&&k1-k2>-0.0000001)||(y1==y2&&y3==y4)) printf("2\n");
        else if((k1*k2+1.0>-0.0000001&&k1*k2+1.0<0.0000001)||(y1==y2&&x3==x4)||(y3==y4&&x1==x2)) printf("1\n");
        else printf("0\n");
    }
}

#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<complex>
using namespace std;
typedef complex<double> qua;
int main()
{
    int t;
    double x1,x2,x3,x4,y1,y2,y3,y4;
    for(scanf("%d",&t);t;--t)
    {
        scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3,&x4,&y4);
        qua ans=qua(x2-x1,y2-y1)*qua(x4-x3,y3-y4);
        if(ans.imag()==0) printf("2\n");
        else if(ans.real()==0) printf("1\n");
        else printf("0\n");
    }
    return 0;
}