POJ 1269 计算几何 题解

Intersecting Lines
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 15336 Accepted: 6713

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

[解题报告]
题目大意:现有一个坐标轴,每次输入两条直线(以四个点的形式给出),询问这两条线段是
重合(LNNE),平行(NONE),还是有交点(并输出交点的坐标)。
思路的话,就分别判断这三种情况。
首先如果重合的话,如果p1,p2,p3共线则mul(p1,p2,p3)=0,同理如果p1,p2,p4共线则mul(p1,p2,p4)=0。(注意程序中要写成fabs(…) eps)。
如果平行的话,则(p2.x-p1.x)* (p4.y-p3.y)==(p2.y-p1.y)*(p4.x-p3.x),没有什么好说的。
如果有交点的话,如何求交点?交点肯定在两条直线上,假设交点坐标是p0.x,P0.y,那么就有:

(p1-p0)*(p2-p0)=0
(p3-p0)*(p4-p0)=0

展开之后用解二元一次方程的公式就可以了。
(在poj上交的时候要用C++,不要用G++)

代码如下:

#includecstdio
#includecstring
#includealgorithm
#includecmath
using namespace std;
#define eps 1e-8

int n;
struct Point
{
    double x,y;
}p1,p2,p3,p4;
double mul(Point p1,Point p2,Point p3)
{
    return (p2.x-p1.x)*(p3.y-p1.y)-(p3.x-p1.x)*(p2.y-p1.y);
}
int main()
{
    scanf("%d",n);
    printf("INTERSECTING LINES OUTPUT\n");
    while(n--)
    {
        scanf("%lf%lf%lf%lf%lf%lf%lf%lf",p1.x,p1.y,p2.x,p2.y,p3.x,p3.y,p4.x,p4.y);
        if(fabs(mul(p1,p2,p3))epsfabs(mul(p1,p2,p4))eps)
        printf("LINE\n");
        else if((p2.x-p1.x)*(p4.y-p3.y)==(p2.y-p1.y)*(p4.x-p3.x))
        printf("NONE\n");
        else
        {
            double a1=p1.y-p2.y;
            double b1=p2.x-p1.x;
            double c1=p1.x*p2.y-p2.x*p1.y;
            double a2=p3.y-p4.y;
            double b2=p4.x-p3.x;
            double c2=p3.x*p4.y-p4.x*p3.y;
            double x=(b1*c2-b2*c1)/(a1*b2-a2*b1);
            double y=(a2*c1-a1*c2)/(a1*b2-a2*b1);
            printf("POINT %.2lf %.2lf\n",x,y);
        }
    }
    printf("END OF OUTPUT\n");
    return 0;
}
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