poj1269 Intersecting Lines(计算几何--两条直线的交点)

Intersecting Lines

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 11307		Accepted: 5110

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.

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

题目大意:在平面直角坐标系中，两点确定一条直线，题目给出4个点，前面2点和后面2点一共确定两条直线，求出这两条直线的交点。
思路:由于两条直线斜率不一定存在，所以方程选择一般式，即ax+by+c = 0,这样比较严谨，两条直线同样存在很多种情况，比如两条直线存在平行，相交，重合3种情况，分情况讨论。代码如下:

#include<iostream>
#include<cstdio>
using namespace std;
typedef struct
{
int x,y;
}Point;
typedef struct
{
int a,b,c;
}Line;
int line(int x1,int y1,int x2,int y2)   //向量(x1,y1),(x2,y2);
{
return x1 * y2 - y1 * x2;
}
Line lineform(int x1,int y1,int x2,int y2)   //ax + by + c = 0;
{
Line temp;
temp.a = y2 - y1;
temp.b = x1 - x2;
temp.c = x2 * y1 - x1 * y2;
return temp;
}
double x,y;
void lineintersect(Line l1,Line l2) //求两条直线的交点
{
double d = l1.a * l2.b - l2.a * l1.b;
x = (l2.c * l1.b - l1.c * l2.b) / d;
y = (l2.a * l1.c - l1.a * l2.c) / d;
}
int main()
{
int n;
Point p1,p2,p3,p4;
Line l1,l2;
//freopen("1.in","r",stdin);
cin>>n;
cout<<"INTERSECTING LINES OUTPUT"<<endl;
while(n--)
{
cin>>p1.x>>p1.y>>p2.x>>p2.y>>p3.x>>p3.y>>p4.x>>p4.y;
if(line(p2.x - p1.x,p2.y - p1.y,p3.x - p1.x,p3.y - p1.y) == 0 && line(p2.x - p1.x,p2.y - p1.y,p4.x - p1.x,p4.y - p1.y) == 0)
{
cout<<"LINE"<<endl; //满足共线的条件
}
else if(line(p2.x - p1.x,p2.y - p1.y,p4.x - p3.x,p4.y - p3.y) == 0)
{
cout<<"NONE"<<endl;//满足平行的条件
}
else
{
l1 = lineform(p1.x,p1.y,p2.x,p2.y);//求两条直线的一般式
l2 = lineform(p3.x,p3.y,p4.x,p4.y);
lineintersect(l1,l2);//求两条直线的交点
printf("POINT %.2lf %.2lf\n",x,y);
}
}
}