HDU 1086 - You can Solve a Geometry Problem too

判断线段两两相交的个数

打模板熟练程度++；

 

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cmath>
 4 using namespace std;
 5 const double EPS = 1e-10;
 6 struct Point 
 7 {
 8     double x,y;
 9     Point(){}
10     Point(double x1,double y1):x(x1),y(y1) {}
11 }a[105],b[105];
12 typedef Point Vect;
13 Vect operator - (Vect A, Vect B) 
14 { 
15     return Vect(A.x - B.x, A.y - B.y); 
16 }
17 int dcmp(double x) 
18 { 
19     return fabs(x) < EPS ? 0 : (x < 0? -1: 1); 
20 }
21 double Cross(Vect A, Vect B) 
22 { 
23     return A.x * B.y - A.y * B.x; 
24 }
25 double Dot(Vect A, Vect B) 
26 { 
27     return A.x * B.x + A.y * B.y; 
28 }
29 bool OnSegment(Point p, Point a1, Point a2)
30 {
31     return dcmp(Cross(a1 - p, a2 - p)) == 0 && dcmp( Dot(a1 - p, a2 - p) ) <= 0;
32 }
33 
34 bool SegmentIntersection(Point a1, Point a2, Point b1, Point b2)
35 {
36     double c1 = Cross(a2 - a1, b1 - a1);
37     double c2 = Cross(a2 - a1, b2 - a1);
38     double c3 = Cross(b2 - b1, a1 - b1);
39     double c4 = Cross(b2 - b1, a2 - b1);
40     if(dcmp(c1) * dcmp(c2) < 0 && dcmp(c3) * dcmp(c4) < 0) return 1;
41     else if(OnSegment(b1, a1, a2) ) return 1;
42     else if(OnSegment(b2, a1, a2) ) return 1;
43     else if(OnSegment(a1, b1, b2) ) return 1;
44     else if(OnSegment(a2, b1, b2) ) return 1;
45     else return 0;
46 }
47 
48 int main()
49 {
50     int n;
51     while(~scanf("%d",&n) && n)
52     {
53         for(int i = 1; i <= n; i++)
54             scanf("%lf%lf%lf%lf", &a[i].x, &a[i].y, &b[i].x, &b[i].y);
55         int ans = 0;
56         for(int i = 1; i <= n; i++)
57             for(int j = i + 1; j <= n; j++)
58                 if(SegmentIntersection(a[i], b[i], a[j], b[j]))
59                     ans++;
60         printf("%d\n", ans);
61     }
62 }

 

posted @ 2016-07-27 22:30  nicetomeetu  阅读(150)  评论(0)  编辑  收藏  举报