題干：

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

題目大意：

? ?T組樣例，每組八個值，前四個值描述一條直線上的兩個點，剩下的同理。

解題報告：

? 套個模板就可以了。

AC代碼：

#include<iostream> #include<algorithm> #include<queue> #include<cstdio> #include<map> #include<vector> #include<set> #include<string> #include<cmath> #include<cstring> #define ll long long #define pb push_back #define pm make_pair #define fi first #define se second using namespace std; const double eps = 1e-8; int sgn(double x) {if(fabs(x) < eps)return 0;if(x < 0) return -1;return 1; } struct Point {double x,y;Point(){}Point(double x,double y):x(x),y(y){}Point operator -(const Point &b)const {return Point(x - b.x,y - b.y);}double operator ^(const Point &b)const {return x*b.y - y*b.x;}double operator *(const Point &b)const {return x*b.x + y*b.y;} }; struct Line {Point s,e;Line(){}Line(Point s,Point e):s(s),e(e){}pair<Point,int> operator &(const Line &b)const {Point res = s;if(sgn((s-e)^(b.s-b.e)) == 0){if(sgn((b.s-s)^(b.e-s)) == 0)return make_pair(res,0);//兩直線重合else return make_pair(res,1);//兩直線平行}double t = ((s-b.s)^(b.s-b.e))/((s-e)^(b.s-b.e));res.x += (e.x - s.x)*t;res.y += (e.y - s.y)*t;return make_pair(res,2);//有交點} }; double xmult(Point p0,Point p1,Point p2) { //p0p1 X p0p2return (p1-p0)^(p2-p0); } bool Seg_inter_line(Line l1,Line l2) { //判斷直線l1和線段l2是否相交，注意是直線！！如果是線段，還需要判斷另一組！！return sgn(xmult(l2.s,l1.s,l1.e))*sgn(xmult(l2.e,l1.s,l1.e)) <= 0; } double dist(Point a,Point b) {return sqrt((b - a)*(b - a)); } int main() {int t;cin>>t;double x1,x2,x3,x4,y1,y2,y3,y4;puts("INTERSECTING LINES OUTPUT");while(t--) {scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3,&x4,&y4);Line l1 = Line(Point(x1,y1),Point(x2,y2));Line l2 = Line(Point(x3,y3),Point(x4,y4));pair<Point,int> p = l1&l2; if(p.se == 1) puts("NONE");else if (p.se == 0) puts("LINE");else printf("POINT %.2f %.2f\n",p.fi.x,p.fi.y);}puts("END OF OUTPUT");return 0 ; }

?

總結(jié)

以上是生活随笔為你收集整理的【POJ - 1269 】Intersecting Lines （计算几何，直线间的位置关系）的全部內(nèi)容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網(wǎng)站內(nèi)容還不錯，歡迎將生活随笔推薦給好友。