POJ1269-Intersecting Lines（判断两条直线平行、重合或者相交）
Intersecting Lines
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 8789 | Accepted: 3965 |
Description
We all know that a pair of distinct points on a plane defines a line and that a pair of lines on a plane will intersect in one of three ways: 1) no intersection because they are parallel, 2) intersect in a line because they are on top of one another (i.e. they are the same line), 3) intersect in a point. In this problem you will use your algebraic knowledge to create a program that determines how and where two lines intersect.
Your program will repeatedly read in four points that define two lines in the x-y plane and determine how and where the lines intersect. All numbers required by this problem will be reasonable, say between -1000 and 1000.
Your program will repeatedly read in four points that define two lines in the x-y plane and determine how and where the lines intersect. All numbers required by this problem will be reasonable, say between -1000 and 1000.
Input
The first line contains an integer N between 1 and 10 describing how many pairs of lines are represented. The next N lines will each contain eight integers. These integers represent the coordinates of four points on the plane in the order x1y1x2y2x3y3x4y4. Thus each of these input lines represents two lines on the plane: the line through (x1,y1) and (x2,y2) and the line through (x3,y3) and (x4,y4). The point (x1,y1) is always distinct from (x2,y2). Likewise with (x3,y3) and (x4,y4).
Output
There should be N+2 lines of output. The first line of output should read INTERSECTING LINES OUTPUT. There will then be one line of output for each pair of planar lines represented by a line of input, describing how the lines intersect: none, line, or point. If the intersection is a point then your program should output the x and y coordinates of the point, correct to two decimal places. The final line of output should read "END OF OUTPUT".
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
这题大概意思就是在平面直角坐标给你两对（四个）点坐标A(x1,y1)和B(x2,y2),C(x3,y3)和D(x4,y4)，让你计算直线A->B和直线C->D是平行(NONE)、重合(LINE)还是相交与一个点(POINT)如果相交就计算出相交点的坐标
这题方法纯粹是用数学知识，所以我们要根据四个点算出两条直线的斜率(K)和b，然后就判断
AC代码：
#include
#include
#include
using namespace std;
double x1,x2,x3,x4,y1,y2,y3,y4,k1,k2,b1,b2,x,y;
int main()
{
int n,i;
while(cin>>n)
{
cout<<"INTERSECTING LINES OUTPUT"<>x1>>y1>>x2>>y2>>x3>>y3>>x4>>y4;
if(x1!=x2&&x3!=x4)
{
k1=(y1-y2)/(x1-x2);//直线A->B的斜率
k2=(y3-y4)/(x3-x4);//直线C->D的斜率
b1=y1-(k1*x1);//直线A->B的b
b2=y3-(k2*x3);//直线C->D的b
if(k1==k2)//如果相等，表明不是平行就是重合
{
if((x3*k1+b1)==y3)//如果直线C->D上点在直线A->B上表明两条直线重合了
cout<<"LINE"<B和C->D的两个点横坐标相等
{
if(x1==x3)//如果四个点都在同一条直线上
cout<<"LINE"<