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
Source
题目类型：直线相交
算法分析：对于每对直线，依次使用叉积判断是否是重合还是相交，然后再判断其中一条线矢量端点是否在另一条直线矢量端点为对角线的矩形内。若判断为相交，则使用矢量计算出交点
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 |
/************************************************* Author :supermaker Created Time :2016/1/24 14:50:18 File Location :C:\Users\abcd\Desktop\TheEternalPoet **************************************************/ #pragma comment(linker, "/STACK:102400000,102400000") #include <set> #include <bitset> #include <list> #include <map> #include <stack> #include <queue> #include <deque> #include <string> #include <vector> #include <ios> #include <iostream> #include <fstream> #include <sstream> #include <iomanip> #include <algorithm> #include <utility> #include <complex> #include <numeric> #include <functional> #include <cmath> #include <ctime> #include <climits> #include <cstdarg> #include <cstdio> #include <cstdlib> #include <cstring> #include <cctype> #include <cassert> using namespace std; #define CFF freopen ("aaa.txt", "r", stdin) #define CPPFF ifstream cin ("aaa.txt") #define DB(ccc) cout << #ccc << " = " << ccc << endl #define PB push_back #define MP(A, B) make_pair(A, B) typedef long long LL; typedef unsigned long long ULL; typedef double DB; typedef pair <int, int> PII; typedef pair <int, bool> PIB; const int INF = 0x7F7F7F7F; const int MOD = 1e9 + 7; const double EPS = 1e-10; const double PI = 2 * acos (0.0); const int maxn = 1e5 + 6666; int sgn (double val) { if (fabs (val) < EPS) return 0; else if (val < 0) return -1; return 1; } struct Point { double x, y; Point () {} Point (double xx, double yy) : x (xx), y (yy) {} Point operator - (const Point &a) const { return Point (x - a.x, y - a.y); } double operator ^ (const Point &a) const { return x * a.y - y * a.x; } double operator * (const Point &a) const { return x * a.x + y * a.y; } }; struct Line { Point s, e; Line () {} Line (Point ss, Point ee) : s (ss), e (ee) {} pair <int, Point> operator & (const Line &a) const { Point res = s; if (sgn ((s - e) ^ (a.s - a.e)) == 0) { if (sgn ((s - a.e) ^ (a.s - a.e)) == 0) return make_pair (0, res); else return make_pair (1, res); } double t = ((s - a.s) ^ (a.s - a.e)) / ((s - e) ^ (a.s - a.e)); res.x += (e.x - s.x) * t; res.y += (e.y - s.y) * t; return make_pair (2, res); } }; int main() { //CFF; //CPPFF; int n; while (scanf ("%d", &n) != EOF) { double x1, y1, x2, y2, x3, y3, x4, y4; puts ("INTERSECTING LINES OUTPUT"); for (int i = 1; i <= n; i++) { scanf ("%lf%lf%lf%lf%lf%lf%lf%lf", &x1, &y1, &x2, &y2, &x3, &y3, &x4, &y4); Line l1 (Point (x1, y1), Point (x2, y2)); Line l2 (Point (x3, y3), Point (x4, y4)); pair <int, Point> tt = l1 & l2; if (tt.first == 0) puts ("LINE"); else if (tt.first == 1) puts ("NONE"); else printf ("POINT %.2lf %.2lf\n", tt.second.x, tt.second.y); } puts ("END OF OUTPUT"); } return 0; } |
- « 上一篇：poj2398
- poj3160：下一篇 »