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 }
hdu 1086 You can Solve a Geometry Problem too (几何),布布扣,bubuko.com