You can Solve a Geometry Problem too
Time
Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K
(Java/Others)
Total Submission(s): 2734 Accepted Submission(s):
1284
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
详细看 《计算几何—基础》
1 #include<iostream>
2 using namespace std;
3 double p[100][2],q[100][2];
4 double direction( double p[] , double q[] , double r[])
5 {
6 return ( (r[0] - p[0]) * (q[1] - p[1]) - (r[1] - p[1]) * (q[0] - p[0]));
7 }
8 bool onsegment(double p[],double q[], double r[])
9 {
10 if(((r[0] - p[0])*(r[0] - q[0]) <= 0) && ((r[1] - p[1]) * (r[1] - q[1]) <= 0))
11 return true;
12 else return false;
13 }
14 bool judge(int i, int j)
15 {
16 double d1 , d2 , d3 , d4 ;
17 d1 = direction(p[i],q[i],p[j]);
18 d2 = direction(p[i],q[i],q[j]);
19 d3 = direction(p[j],q[j],p[i]);
20 d4 = direction(p[j],q[j],q[i]);
21 if( (d1 * d2 < 0) && (d3 * d4 < 0))
22 return true;
23 else if( d1 == 0 && onsegment(p[i],q[i],p[j]) == 1)
24 return true;
25 else if( d2 == 0 && onsegment(p[i],q[i],q[j]) == 1)
26 return true;
27 else if( d3 == 0 && onsegment(p[j],q[j],p[i]) == 1)
28 return true;
29 else if( d4 == 0 && onsegment(p[j],q[j],q[i]) == 1)
30 return true;
31 else return false;
32 }
33 int main(){
34 int n, i , j , count;
35 while( scanf( "%d" , &n ) , n)
36 {
37 for( i = 0 ; i < n ; ++ i )
38 {
39 scanf( "%lf %lf %lf %lf", &p[i][0] , &p[i][1] , &q[i][0] , &q[i][1] ) ;
40 }
41 for( i = 0 , count = 0 ; i < n ; i ++ )
42 for( j = i + 1 ; j < n ; j ++ )
43 if( judge( i , j ) == 1 )
44 ++count;
45 printf("%d\n",count);
46 }
47 return 0;
48 }