POJ 1269 Intersecting Lines 计算几何

                                       Intersecting Lines

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

计算几何基础

AC Code: 

#include <iostream>
#include <queue>
#include <stack>
#include <cstdio>
#include <string>
#include <cstring>
#include <cmath>
#include <vector>
#include <algorithm>
#include <map>
#include <set>
#define Mod 1000000007
#define INF 0x3f3f3f3f
#define eps 1e-8
typedef long long ll;
using namespace std;
static const int MAX_N = 100005;
struct Point{
	double x;
	double y;
}p1, p2, p3, p4;
double direction(Point p1, Point p2, Point p3) {	//p1,p2与p1,p3的叉乘
	return (p1.x - p2.x) * (p1.y - p3.y) - (p1.x - p3.x) * (p1.y - p2.y);
}
int main() {
	int T;
	scanf("%d", &T);
	printf("INTERSECTING LINES OUTPUT\n");
	while (T--) {
		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(direction(p1, p2, p3) < eps && fabs(direction(p1, p2, p4)) < eps)) {	//共线平行
			printf("LINE\n");
		}
		else if ((p2.x - p1.x) * (p4.y - p3.y) == (p2.y - p1.y) * (p4.x - p3.x)) {	//不共线平行
			printf("NONE\n");
		}
		else {	//相交
			double a1 = p1.y - p2.y;
			double b1 = p2.x - p1.x;
			double c1 = p1.x * p2.y - p1.y * p2.x;
			double a2 = p3.y - p4.y;
			double b2 = p4.x - p3.x;
			double c2 = p3.x * p4.y - p3.y * p4.x;
			double x = (b1 * c2 - b2 * c1) / (a1 * b2 - a2 * b1);
			double y = (a2 * c1 - a1 * c2) / (a1 * b2 - a2 * b1);	//求交点
			printf("POINT %.2f %.2f\n", x, y);
		}
	}
	printf("END OF OUTPUT\n");
	return 0;
}