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Intersecting Lines
Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 12421 Accepted: 5548 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 5Sample Output
INTERSECTING LINES OUTPUT POINT 2.00 2.00 NONE LINE POINT 2.00 5.00 POINT 1.07 2.20 END OF OUTPUTSource
题意：给定 1 - 10组直线，判断每组直线的关系，若相交 输出交点坐标，保留两位小数;若平行，输出‘NONE’;若重合，输出‘LINE’;
输出格式详见标准输出。
1 #include <iostream> 2 #include <cstdio> 3 #include <cmath> 4 #include <cstdlib> 5 #include <cstring> 6 #include <math.h> 7 #include <algorithm> 8 #include <cctype> 9 #include <string> 10 #include <map> 11 #include <set> 12 #define ll long long 13 using namespace std; 14 const double eps = 1e-8; 15 int sgn(double x) 16 { 17 if(fabs(x) < eps)return 0; 18 if(x < 0) return -1; 19 else return 1; 20 } 21 struct Point 22 { 23 double x,y; 24 Point(){} 25 Point(double _x,double _y) 26 { 27 x = _x;y = _y; 28 } 29 Point operator -(const Point &b)const 30 { 31 return Point(x - b.x,y - b.y); 32 } 33 double operator ^(const Point &b)const 34 { 35 return x*b.y - y*b.x; 36 } 37 double operator *(const Point &b)const 38 { 39 return x*b.x + y*b.y; 40 } 41 }; 42 43 struct Line 44 { 45 Point s,e; 46 Line(){} 47 Line(Point _s,Point _e) 48 { 49 s = _s;e = _e; 50 } 51 pair<Point,int> operator &(const Line &b)const 52 { 53 Point res = s; 54 if(sgn((s-e)^(b.s-b.e)) == 0) 55 { 56 if(sgn((b.s-s)^(b.e-s)) == 0) 57 return make_pair(res,0);//两直线重合 58 else return make_pair(res,1);//两直线平行 59 } 60 double t = ((s-b.s)^(b.s-b.e))/((s-e)^(b.s-b.e)); 61 res.x += (e.x - s.x)*t; 62 res.y += (e.y - s.y)*t; 63 return make_pair(res,2);//有交点 64 } 65 }; 66 67 int main(void) 68 { 69 int t; 70 double x1,x2,x3,x4,y1,y2,y3,y4; 71 scanf("%d",&t); 72 printf("INTERSECTING LINES OUTPUT\n"); 73 while(t--) 74 { 75 scanf("%lf %lf %lf %lf %lf %lf %lf %lf",&x1,&y1,&x2,&y2,&x3,&y3,&x4,&y4); 76 Line l1 = Line( Point(x1,y1) ,Point(x2,y2) ); 77 Line l2 = Line( Point(x3,y3) ,Point(x4,y4) ); 78 pair<Point,int> ans = l1 & l2; 79 if(ans.second == 2) printf("POINT %.2f %.2f\n",ans.first.x,ans.first.y); 80 else if(ans.second == 0) printf("LINE\n"); 81 else printf("NONE\n"); 82 } 83 printf("END OF OUTPUT\n"); 84 85 return 0; 86 }
poj 1269 Intersecting Lines(判断两直线关系，并求交点坐标)
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原文地址：http://www.cnblogs.com/henserlinda/p/4736099.html
