[POJ1269]Intersecting Lines (计算几何)

题目链接：http://poj.org/problem?id=1269

题意：给两条直线，判断这两条直线是平行中的重合还是相交。

一切尽在不言中：

  1 #include <algorithm>
  2 #include <iostream>
  3 #include <iomanip>
  4 #include <cstring>
  5 #include <climits>
  6 #include <complex>
  7 #include <fstream>
  8 #include <cassert>
  9 #include <cstdio>
 10 #include <bitset>
 11 #include <vector>
 12 #include <deque>
 13 #include <queue>
 14 #include <stack>
 15 #include <ctime>
 16 #include <set>
 17 #include <map>
 18 #include <cmath>
 19 
 20 using namespace std;
 21 
 22 const double eps = 1e-8;
 23 
 24 typedef struct Point {
 25     double x;
 26     double y;
 27     Point() {}
 28     Point(double xx, double yy) : x(xx), y(yy) {}
 29     Point operator +(const Point& b) const { return Point(x+b.x, y+b.y); }
 30     Point operator -(const Point& b) const { return Point(x-b.x, y-b.y); }
 31     double operator ^(const Point& b) const { return x * b.y - y * b.x; }
 32     double operator *(const Point& b) const { return x * b.x + y * b.y; }
 33 }Point;
 34 
 35 typedef struct Line {
 36     Point u;
 37     Point v;
 38     Line() {}
 39     Line(Point uu, Point vv) : u(uu), v(vv) {}
 40 }Line;
 41 
 42 int sgn(double x) {
 43     if(fabs(x) < eps) return 0;
 44     if(x < 0) return -1;
 45     return 1;
 46 }
 47 
 48 double xmulti(Point p0, Point p1, Point p2) {
 49     return (p1 - p0) ^ (p2 - p0);
 50 }
 51 
 52 double dist(Point a, Point b) {
 53     return sqrt((b-a)*(b-a));
 54 }
 55 
 56 double intersect(Line l1, Line l2) {
 57     return sgn(xmulti(l2.u, l1.u, l1.v))*sgn(xmulti(l2.v, l1.u, l1.v))<=0;
 58 }
 59 
 60 double ex1, ey1, ex2, ey2;
 61 double ex3, ey3, ex4, ey4;
 62 
 63 pair<Point, int> check(Line l1, Line l2) {
 64     Point ans = l1.u;
 65     if(sgn((l1.u-l1.v)^(l2.u-l2.v)) == 0) {
 66         if(sgn((l2.u-l1.u)^(l2.v-l1.u)) == 0) {
 67             return pair<Point, int>(ans, 0);//chonghe
 68         }
 69         else {            
 70             return pair<Point, int>(ans, 1);//pingxing
 71         }
 72     }
 73     double t = ((l1.u - l2.u) ^ (l2.u - l2.v)) / ((l1.u - l1.v) ^ (l2.u - l2.v));
 74     ans.x += (l1.v.x - l1.u.x) * t;
 75     ans.y += (l1.v.y - l1.u.y) * t;
 76     return pair<Point, int>(ans, 2);//jiaodian ans
 77 }
 78 
 79 int main() {
 80     // freopen("in", "r", stdin);
 81     int T;
 82     scanf("%d", &T);
 83     printf("INTERSECTING LINES OUTPUT\n");
 84     while(T--) {
 85         scanf("%lf %lf %lf %lf", &ex1, &ey1, &ex2, &ey2);
 86         scanf("%lf %lf %lf %lf", &ex3, &ey3, &ex4, &ey4);
 87         Line l1 = Line(Point(ex1, ey1), Point(ex2, ey2));
 88         Line l2 = Line(Point(ex3, ey3), Point(ex4, ey4));
 89         pair<Point, int> ans = check(l1, l2);
 90         if(ans.second == 0) {
 91             printf("LINE\n");
 92         }
 93         else if(ans.second == 1) {
 94             printf("NONE\n");
 95         }
 96         else {
 97             printf("POINT %.2f %.2f\n", ans.first.x, ans.first.y);
 98         }
 99     }
100     printf("END OF OUTPUT\n");
101     return 0;
102 }

 

posted @ 2016-02-21 19:45  Kirai  阅读(242)  评论(0)  编辑  收藏  举报