POJ 1269 Intersecting Lines

Intersecting Lines

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 7341		Accepted: 3381

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

Source

Mid-Atlantic 1996

 

 

题目意思就是让你判断两条直线是相交，平行，还是 共线

相交输出交点。

输出交点时被卡了下，不想用最传统的方法求：

后来发现可以利用叉积，求出面积比，再得出交点坐标

 

#include<stdio.h>
#include<math.h>
#define eps 1e-8
double multi(double x1,double y1,double x2,double y2,double x3,double y3) {
    return (x2-x1)*(y3-y1) - (x3-x1)*(y2-y1) ;
}
double Area(double x1,double y1,double x2,double y2,double x3,double y3){
    return (x2-x1)*(y3-y1) - (x3-x1)*(y2-y1) ;
}
int main(){
    int T;
    scanf("%d",&T);
    puts("INTERSECTING LINES OUTPUT") ;
    while(T--){
        double x1,x2,x3,x4,y1,y2,y3,y4 ;
        scanf("%lf %lf %lf %lf",&x1,&y1,&x2,&y2) ;
        scanf("%lf %lf %lf %lf",&x3,&y3,&x4,&y4) ;
        double flag = (x2-x1)*(y4-y3) - (y2-y1)*(x4-x3) ;
        if(fabs(flag) <= eps) {
            if(fabs((x2-x1)*(y3-y1) - (x3-x1)*(y2-y1)) <= eps)
                puts("LINE") ;
            else
                puts("NONE") ;
        }else{
            double x,y ;
            double s1 = Area(x1,y1,x3,y3,x4,y4) ;
            double s2 = -Area(x2,y2,x3,y3,x4,y4) ;
            double temp = s1 / (s1+s2);
            x = x1 + (x2 - x1) * temp ;
            y = y1 + (y2 - y1) * temp ;
            printf("POINT %.2lf %.2lf\n",x,y) ;
        }
    }
    puts("END OF OUTPUT") ;
    return 0;
}