ACM-计算几何之Intersecting Lines——poj1269

Intersecting Lines

题目：http://poj.org/problem?id=1269

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

这道题说是计算几何，其实用解析几何做就可以了。

给你两条直线，每条直线给两个点（两点确定一条直线嘛~）

判断这两条直线是：相交or相离or重合

嗯，就是给两个点，求直线方程~\(≧▽≦)/~啦啦啦

首先是判断两条直线斜率是否都存在，如果斜率都不存在则平行或重合，判断两条直线的横坐标是否相等，相等则重合，不相等则平行。（垂直于x轴）

再处理只有一个斜率存在，只有一个斜率存在，那肯定就是相交，因为平行和重合的之前判断过了。

交点可以确定了，就是斜率不存在那条直线的横坐标，显然直接通过方程求纵坐标即可。

最后处理两条直线斜率都存在。

那就判断斜率是否相等，再判断斜距是否相等：

斜率相等：

斜距相等→重合

斜距不等→平行

斜率不相等，相交。

知道两条直线的方程，求交点坐标，应该不用说了。

注意一下输出格式问题，OK，AC咯~

#include <stdio.h>struct Point{    double x,y;};struct Line{    Point a,b;};// 为0则平行，为1则同一条直线，为2则相交于一点.int ispos;// 交点坐标及斜率、斜距值double p_x,p_y,k1,k2,b1,b2;double abs(double x){    return x<0?-x:x;}bool find_k(Line l,int num){    if(l.a.x==l.b.x)    return false;    if(num==1)    {       k1=(l.b.y-l.a.y)/(l.b.x-l.a.x);       b1=l.a.y-k1*l.a.x;    }    else    {        k2=(l.b.y-l.a.y)/(l.b.x-l.a.x);        b2=l.a.y-k2*l.a.x;    }    return true;}void judge(Line m,Line n){    // 两条直线斜率K都不存在，即两直线都垂直于x轴    // 这里用 & ，让两个函数务必都执行，这样k1,k2,b1,b2都算出来了，可以直接调用了    if(!find_k(m,1) & !find_k(n,2))    {        if(m.a.x==n.a.x)    ispos=1;        return;    }    // 有任何一条直线斜率K不存在    if(!find_k(m,1))    {        p_x=m.a.x;        p_y=k2*p_x+b2;        ispos=2;        return;    }    else if(!find_k(n,2))    {        p_x=n.a.x;        p_y=k1*p_x+b1;        ispos=2;        return;    }    // 两条直线斜率k都存在    if(k1==k2)    {        if(b1==b2)  ispos=1;        return;    }    p_x=(b2-b1)/(k1-k2);    p_y=k1*p_x+b1;    ispos=2;    return;}int main(){    int n;    Line l1,l2;    scanf("%d",&n);    printf("INTERSECTING LINES OUTPUT\n");    while(n--)    {        p_x=p_y=k1=k2=b1=b2=0;        scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&l1.a.x,&l1.a.y,&l1.b.x,&l1.b.y,&l2.a.x,&l2.a.y,&l2.b.x,&l2.b.y);        ispos=0;        judge(l1,l2);        if(!ispos)  printf("NONE\n");        else if(ispos==1)   printf("LINE\n");        else    printf("POINT %.2lf %.2lf\n",p_x,p_y);    }    printf("END OF OUTPUT\n");    return 0;}