模板 线段相交

【模板】线段相交

 1 #include <stdio.h>
 2 #include <iostream>
 3 #include <math.h>
 4 #include <algorithm>
 5 using namespace std;
 6 struct point
 7 {
 8     double x,y;
 9 };
10 point a[105][2];//a[i][0]代表第i条线段的头，a[i][1]代表尾
11 double fan(double x,double y)
12 {
13     return x>y?x:y;
14 }
15
16 double fin(double c,double d)
17 {
18     return c<d?c:d;
19 }
20
21 double cnt(point a,point b)
22 {
23     return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
24 }
25
26 int is(point a,point b,point c,point d)
27 {
28     if(a.x==b.x&&c.x==d.x)
29     {
30         return 0;
31     }
32     if(a.x==b.x&&c.x!=d.x)
33     {
34         double m1=a.x;
35         double m2=(a.x-c.x)*(d.y-c.y)/(d.x-c.x)+c.y;
36         if(m1<=fan(a.x,b.x)&&m1>=fin(a.x,b.x)&&m2>=fin(a.y,b.y)&&m2<=fan(a.y,b.y)&&m1<=fan(c.x,d.x)&&m1>=fin(c.x,d.x)&&m2>=fin(c.y,d.y)&&m2<=fan(c.y,d.y))
37             return 1;
38     }
39     if(c.x==d.x&&a.x!=b.x)
40     {
41         double m1=c.x;
42         double m2=a.y+(b.y-a.y)*(c.x-a.x)/(b.x-a.x);
43         if(m1<=fan(a.x,b.x)&&m1>=fin(a.x,b.x)&&m2>=fin(a.y,b.y)&&m2<=fan(a.y,b.y)&&m1<=fan(c.x,d.x)&&m1>=fin(c.x,d.x)&&m2>=fin(c.y,d.y)&&m2<=fan(c.y,d.y))
44             return 1;
45     }
46     double k1=(b.y-a.y)/(b.x-a.x);
47     double k2=(d.y-c.y)/(d.x-c.x);
48     double m1,m2,x,y;
49     if(k1==k2)  return 0;
50     else
51     {
52         m1=a.y-k1*a.x;
53         m2=c.y-k2*c.x;
54         x=(m1-m2)/(k2-k1);
55         y=k1*x+m1;
56         if(x<=fan(a.x,b.x)&&x>=fin(a.x,b.x)&&y>=fin(a.y,b.y)&&y<=fan(a.y,b.y)&&x<=fan(c.x,d.x)&&x>=fin(c.x,d.x)&&y>=fin(c.y,d.y)&&y<=fan(c.y,d.y))
57             return 1;
58     }
59     return 0;
60 }
61
62 int main()
63 {
64     int cas = 1;
65     int n,i,j;
66     while(~scanf("%d",&n),n)
67     {
68         int cnt = 0;
69         for(i = 0;i<n;i++)
70         scanf("%lf%lf%lf%lf",&a[i][0].x,&a[i][0].y,&a[i][1].x,&a[i][1].y);//线段的首尾坐标
71         for(i = 0;i<n;i++)
72         {
73             for(j = i+1;j<n;j++)
74           {
75 if(is(a[i][0],a[i][1],a[j][0],a[j][1]))
76                 cnt++;
77             }
78         }
79         printf("%d\n",cnt);
80     }
81     return 0;
82 }