Intersecting Lines

Intersecting Lines

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 5339		Accepted: 2526

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

北大的数据太水了。能ac .在我们学校oj.。比赛时做了好久。。一直wa..一直用斜率做的。。

最后精度有问题。加fabs同样也不行。痛苦。。今天tao用向量1a,羡慕嫉妒恨啊。。

也用向量做了一次。。终于AC了。。

推公式好痛苦。。急需耐心。。

#include <stdio.h>
#include <string.h>
#include <stdio.h>

void debug( )
{   
    #ifdef P
    freopen("1.in","r",stdin);
    freopen("2.out","w",stdout);
    #endif
}

int main( )
{
    int N;
    debug( );
    scanf("%d",&N);
    puts("INTERSECTING LINES OUTPUT");
    while (N--)
    {
        double x1,y1,x2,y2,x3,y3,x4,y4, p1,p2,q1,q2,b1,b2,v1,v2;
        scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&x1, &y1, &x2, &y2,&x3,&y3,&x4,&y4);
        p1 = x2 - x1;
        p2 = y2 - y1;
        q1 = x4 - x3;
        q2 = y4 -y3;
        if ( p1 * q2 == p2 * q1 )
        {
            
            if ( p1 == 0 )
            {
                    if ( x1 == x3)
                        puts("LINE");
                    else
                        puts("NONE");
            }
            else if (p2 == 0)
            {
                     if (y1 == y3)
                         puts("LINE");
                     else
                         puts("NONE");
            }
            else
            {  
                   if ( ( y1 - p2 / p1 * x1 ) == (y3 - q2 / q1 * x3) )
                           puts("LINE");
                   else
                           puts("NONE");
            }
        }
        else
        {
            if ( p1 * q1 + q2 * p2 == 0 )
             {
                    if( p1 == 0)
                        {
                            printf("POINT %.2lf %.2lf\n", x1, y3);
                        }
                    else  if (q1 == 0)
                        {
                            printf("POINT %.2lf %.2lf\n", x3, y1);
                        }
                    else 
                        {
                            v1 = (p1 * q1 * (y3 -y1) + p2 * q1 * x1 - p1 *q2 * x3) / ( p2 * q1 - p1 * q2);
                            v2 = (q2 * p2 * (x3 -  x1) + p1 * q2 * y1 - p2 * q1 * y3) / ( p1 * q2 - q1 * p2);
                            printf("POINT %.2lf %.2lf\n", v1, v2);
                        }
                             
                         
                        
            }
            else
                 {         
                            v1 = (p1 * q1 * (y3 -y1) + p2 * q1 * x1 - p1 *q2 * x3) / ( p2 * q1 - p1 * q2);
                            v2 = (q2 * p2 * (x3 -  x1) + p1 * q2 * y1 - p2 * q1 * y3) / ( p1 * q2 - q1 * p2);
                            printf("POINT %.2lf %.2lf\n", v1, v2);
                 }
                  
        }
    }
    printf("END OF OUTPUT\n");
    return 0;
}