POJ 1269 Intersecting Lines

题目大意：

给定n组数据，每组2条直线，求直线关系，关系有如下

1.重合

2.平行

3.相交于一点

输入：

有n行，每行8个数，表示第一条直线的两点和第二条直线的两点

输出：

格式参照输出样例，对于相交，输出交点

Sample Input

5
0 0 4 4 0 4 4 0
5 0 7 6 1 0 2 3
5 0 7 6 3 -6 4 -3
2 0 2 27 1 5 18 5
0 3 4 0 1 2 2 5

Sample Output

INTERSECTING LINES OUTPUT
POINT 2.00 2.00
NONE
LINE
POINT 2.00 5.00
POINT 1.07 2.20
END OF OUTPUT
题解：
这题不难，主要是函数计算，只要有基础的一次函数知识就能解出
但最重要的是斜率不存在的情况，要特判

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
using namespace std;
double x1,y11,x2,y2,x3,y3,x4,y4;
int n;
int main()
{int i;
    cin>>n;
    cout<<"INTERSECTING LINES OUTPUT
";
    for (i=1;i<=n;i++)
    {
        cin>>x1>>y11>>x2>>y2>>x3>>y3>>x4>>y4;
        if (x2-x1==0&&x4-x3==0)
        {
            if (x1!=x3)
            cout<<"NONE
";
            else cout<<"LINE
";
        }
        else
        if (x2-x1==0)
        {
            cout<<"POINT ";
             double x=x1;
              printf("%.2lf ",x);
              double y=((y4-y3)*x+(x4*y3-x3*y4))/(x4-x3);
              printf("%.2lf
",y);
        }
        else 
        if (x4-x3==0)
        {
            cout<<"POINT ";
             double x=x3;
              printf("%.2lf ",x);
              double y=((y2-y11)*x+(x2*y11-x1*y2))/(x2-x1);
              printf("%.2lf
",y);
        }
        else
        if ((y2-y11)*(x4-x3)==(y4-y3)*(x2-x1))
        {
            if ((x2-x1)*(x4*y3-x3*y4)==(x4-x3)*(x2*y11-x1*y2))
            {
                cout<<"LINE
";
            }
            else cout<<"NONE
";
        }
         else 
         {
             double fz=-(x2*y11-x1*y2)*(x4-x3)+(x4*y3-x3*y4)*(x2-x1);
             double fm=(y2-y11)*(x4-x3)-(y4-y3)*(x2-x1);
             cout<<"POINT ";
             double x=fz/fm;
              printf("%.2lf ",x);
              double y=(y2-y11)*x/(x2-x1)+(x2*y11-x1*y2)/(x2-x1);
              printf("%.2lf
",y);
         }
    }
    cout<<"END OF OUTPUT";
}