poj 1269(两条直线交点)

Intersecting Lines

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 13481		Accepted: 5997

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

计算几何真的不容易AC。。。模板代码。。函数名自己用百度翻译起的。。

#include <iostream>
#include <cstdio>
#include <string.h>
#include <math.h>
#include <algorithm>

using namespace std;
const double eps = 1e-8;
const int N = 105;
struct Point
{
double x,y;
} ;
struct Line{
Point a,b;
};
///叉积
double mult(Point a, Point b, Point c)
{
return (a.x-c.x)*(b.y-c.y)-(b.x-c.x)*(a.y-c.y);
}
///计算两条直线的交点
Point intersection(Point a,Point b,Point c,Point d){
Point p = a;
double t = ((a.x-c.x)*(c.y-d.y)-(a.y-c.y)*(c.x-d.x))/((a.x-b.x)*(c.y-d.y)-(a.y-b.y)*(c.x-d.x));
p.x +=(b.x-a.x)*t;
p.y +=(b.y-a.y)*t;
return p;
}
double parallel(Point a,Point b,Point c,Point d){ ///判断平行
return fabs((a.x-b.x)*(c.y-d.y)-(a.y-b.y)*(c.x-d.x));
}
int main()
{
int tcase;
scanf("%d",&tcase);
printf("INTERSECTING LINES OUTPUT\n");
while(tcase--){
Line l1,l2;
scanf("%lf%lf%lf%lf",&l1.a.x,&l1.a.y,&l1.b.x,&l1.b.y);
scanf("%lf%lf%lf%lf",&l2.a.x,&l2.a.y,&l2.b.x,&l2.b.y);
if(parallel(l1.a,l1.b,l2.a,l2.b)<eps){ ///两直线平行
if(fabs(mult(l1.b,l2.a,l1.a))<eps){ ///判断共线
printf("LINE\n");
}
else printf("NONE\n");
}
else{
Point p = intersection(l1.a,l1.b,l2.a,l2.b);
printf("POINT %.2lf %.2lf\n",p.x,p.y);
}

}
printf("END OF OUTPUT\n");
return 0;
}