You can Solve a Geometry Problem too
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6425 Accepted Submission(s): 3099
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.
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.
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
Recommend
计算几何：判断两线段是否相交。
很简单的一道题，套上模板之后直接遍历判断每对线段是否相交，相交就计数，最后输出计数就是交点数。这种题的思路就是做两个验证，这两个验证学名叫快速排斥实验和跨立实验，分别有4个判断和2个判断，只有这两个实验都通过才能说这两条线段相交。详见：
判断两线段是否相交模板：
1 struct Point{
2 double x,y;
3 };
4 struct Line{
5 Point p1,p2;
6 };
7 double Max(double a,double b)
8 {
9 return a>b?a:b;
10 }
11 double Min(double a,double b)
12 {
13 return a<b?a:b;
14 }
15 double xmulti(Point p1,Point p2,Point p0)
16 {
17 return (p1.x-p0.x)*(p2.y-p0.y)-(p1.y-p0.y)*(p2.x-p0.x);
18 }
19 bool inter(Line l1,Line l2)
20 {
21 if( Min(l2.p1.x,l2.p2.x)<=Max(l1.p1.x,l1.p2.x) &&
22 Min(l2.p1.y,l2.p2.y)<=Max(l1.p1.y,l1.p2.y) &&
23 Min(l1.p1.x,l1.p2.x)<=Max(l2.p1.x,l2.p2.x) &&
24 Min(l1.p1.y,l1.p2.y)<=Max(l2.p1.y,l2.p2.y) &&
25 xmulti(l1.p1,l2.p2,l2.p1)*xmulti(l1.p2,l2.p2,l2.p1)<=0 &&
26 xmulti(l2.p1,l1.p2,l1.p1)*xmulti(l2.p2,l1.p2,l1.p1)<=0 )
27 return true;
28 else
29 return false;
30 }
本题代码：
1 #include <iostream>
2 using namespace std;
3 struct Point{
4 double x,y;
5 };
6 struct Line{
7 Point p1,p2;
8 };
9 double Max(double a,double b)
10 {
11 return a>b?a:b;
12 }
13 double Min(double a,double b)
14 {
15 return a<b?a:b;
16 }
17 double xmulti(Point p1,Point p2,Point p0)
18 {
19 return (p1.x-p0.x)*(p2.y-p0.y)-(p1.y-p0.y)*(p2.x-p0.x);
20 }
21 bool inter(Line l1,Line l2)
22 {
23 if( Min(l2.p1.x,l2.p2.x)<=Max(l1.p1.x,l1.p2.x) &&
24 Min(l2.p1.y,l2.p2.y)<=Max(l1.p1.y,l1.p2.y) &&
25 Min(l1.p1.x,l1.p2.x)<=Max(l2.p1.x,l2.p2.x) &&
26 Min(l1.p1.y,l1.p2.y)<=Max(l2.p1.y,l2.p2.y) &&
27 xmulti(l1.p1,l2.p2,l2.p1)*xmulti(l1.p2,l2.p2,l2.p1)<=0 &&
28 xmulti(l2.p1,l1.p2,l1.p1)*xmulti(l2.p2,l1.p2,l1.p1)<=0 )
29 return true;
30 else
31 return false;
32 }
33 int main()
34 {
35 int N;
36 Line l[105];
37 while(cin>>N){
38 if(N==0) break;
39 for(int i=1;i<=N;i++)
40 cin>>l[i].p1.x>>l[i].p1.y>>l[i].p2.x>>l[i].p2.y;
41 int c = 0;
42 for(int i=1;i<=N-1;i++)
43 for(int j=i+1;j<=N;j++)
44 if(inter(l[i],l[j]))
45 c++;
46 cout<<c<<endl;
47 }
48 return 0;
49 }
Freecode : www.cnblogs.com/yym2013