POJ 1269(计算几何+直线相交)

问题描述:

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

50 0 4 4 0 4 4 05 0 7 6 1 0 2 35 0 7 6 3 -6 4 -32 0 2 27 1 5 18 50 3 4 0 1 2 2 5

Sample Output

INTERSECTING LINES OUTPUTPOINT 2.00 2.00NONELINEPOINT 2.00 5.00POINT 1.07 2.20END OF OUTPUT

题目题意:题目给我们四个点的坐标，其中前俩点确定一条直线，后俩个点确定一条直线，问这俩条直线的位置关系（1：平行，2：重合 3：相交（并求交点))

题目分析:因为欧几里得几何表示直线方程不全面，所以最好用向量来处理。

俩条直线平行即它们的方向向量平行。

如果俩条直线平行了，那么有可能重合，怎么确定重合了？

  。        。      。                          。         。        。                                    。        。          。

  a          c        d                           c            a         d                                    c           d             a

如果俩条直线重合,只要有一点例如点a（排除相交了，因为他们平行)在直线cd 上那么直线ac重合cd 

a点在直线cd上的位置关系三种，用距离相等可以列出判断表达式

dis(a,c)+dis(c,d)==dis(a,d)||dis(a,c)+dis(a,d)==dis(c,d)||dis(a,d)+dis(c,d)==dis(a,c) 如果成立就重合

如果不满足以上俩种情况就相交求交点。

直线ac 与x轴的夹角angle1,直线cd 与x轴夹角angle2

则ac: (x1+cos(angle1)*t1,y1+sin(angle1)*t1)

   cd:(x2+cos(angle2)*t2,y2+sin(angle2)*t2)

列等式求解即可
代码如下：

#include<iostream>#include<cstdio>#include<cmath>#include<cstring>using namespace std;const double epx=1e-6;struct note{    double x,y;};double get_dis(note a,note b){    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));}bool isparallel(note a,note b,note c,note d)//是否平行{    return (a.x-b.x)*(c.y-d.y)==(c.x-d.x)*(a.y-b.y);}bool check(note a,note c,note d)//是否重合{    double len1=get_dis(a,c);    double len2=get_dis(a,d);    double len3=get_dis(c,d);    return (len1+len3)-len2<epx||(len1+len2)-len3<epx||(len2+len3)-len1<epx;}int main(){    int t;    scanf("%d",&t);    printf("INTERSECTING LINES OUTPUT\n");    while (t--) {        struct note a,b,c,d;        scanf("%lf%lf%lf%lf",&a.x,&a.y,&b.x,&b.y);        scanf("%lf%lf%lf%lf",&c.x,&c.y,&d.x,&d.y);        if (isparallel(a,b,c,d)){            if (check(a,c,d)) {                printf("LINE\n");                continue;            }            else {                printf("NONE\n");                continue;            }        }        double angle1=atan((b.y-a.y)/(b.x-a.x));        double angle2=atan((d.y-c.y)/(d.x-c.x));        double k1,k2,k3,k4;        k1=cos(angle1);        k2=sin(angle1);        k3=cos(angle2);        k4=sin(angle2);        double res=(k4*c.x-k3*c.y+a.y*k3-k4*a.x)/(k1*k4-k3*k2);        printf("POINT %.2f %.2f\n",a.x+k1*res,a.y+k2*res);    }    printf("END OF OUTPUT\n");    return 0;}