hdu 1086 You can Solve a Geometry Problem too (几何)

You can Solve a Geometry Problem too

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6932    Accepted Submission(s): 3350


Problem Description
Many geometry（几何）problems were designed in the ACM/ICPC. And now, I also prepare a geometry problem for this final exam. According to the experience of many ACMers, geometry problems are always much trouble, but this problem is very easy, after all we are now attending an exam, not a contest :)
Give you N (1<=N<=100) segments（线段）, please output the number of all intersections（交点）. You should count repeatedly if M (M>2) segments intersect at the same point.

Note:
You can assume that two segments would not intersect at more than one point. 
 

 

Input
Input contains multiple test cases. Each test case contains a integer N (1=N<=100) in a line first, and then N lines follow. Each line describes one segment with four float values x1, y1, x2, y2 which are coordinates of the segment’s ending. 
A test case starting with 0 terminates the input and this test case is not to be processed.
 

 

Output
For each case, print the number of intersections, and one line one case.
 

 

Sample Input
2
0.00 0.00 1.00 1.00
0.00 1.00 1.00 0.00
3
0.00 0.00 1.00 1.00
0.00 1.00 1.00 0.000
0.00 0.00 1.00 0.00
0
 

 

Sample Output
1
3
 

 

Author
lcy
 

 

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 简单数学几何，求n条线段共有几个交点。

 1 //0MS    240K    1146 B    C++
 2 #include<stdio.h> 
 3 #include<math.h>
 4 struct node{
 5     double x1,y1;
 6     double x2,y2;        
 7 }p[105];
 8 double Max(double a,double b)
 9 {
10     return a>b?a:b;
11 }
12 double Min(double a,double b)
13 {
14     return a<b?a:b;
15 }
16 int judge_in(node a,double x,double y)
17 {
18     if(x>=Min(a.x1,a.x2)&&x<=Max(a.x1,a.x2)&&y>=Min(a.y1,a.y2)&&y<=Max(a.y1,a.y2))
19         return 1;
20     return 0;
21 }
22 int judge(node a,node b)
23 {
24     double k1,k2,b1,b2;
25     if(a.x1==a.x2) k1=0;
26     else k1=(a.y2-a.y1)/(a.x2-a.x1);
27     if(b.x1==b.x2) k2=0;
28     else k2=(b.y2-b.y1)/(b.x2-b.x1);
29     if(k1==k2) return 0;
30     
31     b1=a.y1-k1*a.x1;
32     b2=b.y1-k2*b.x1;
33     
34     double x,y;
35     x=(b2-b1)/(k1-k2);
36     y=k1*x+b1;
37     
38     if(judge_in(a,x,y) && judge_in(b,x,y))  return 1;
39     return 0; 
40 }
41 int main(void)
42 {
43     int n;
44     while(scanf("%d",&n)!=EOF && n)
45     {
46         for(int i=0;i<n;i++)
47             scanf("%lf%lf%lf%lf",&p[i].x1,&p[i].y1,&p[i].x2,&p[i].y2);
48         int cnt=0;
49         for(int i=0;i<n;i++)
50             for(int j=i+1;j<n;j++)
51                 cnt+=judge(p[i],p[j]);
52         printf("%d\n",cnt);
53     }
54     return 0;
55 }

 

posted @ 2014-06-02 10:31  heaventouch  阅读(238)  评论(0)  编辑  收藏  举报