Graphics computing mathematics notes (continuously updated). . .

// hdu1086 find an intersection point number n of segments. . .

//http://acm.hdu.edu.cn/showproblem.php?pid=1086

// The following is a tip translated:

// you N (1 <= N <= 100) a segment (segments), output a number of all intersection points. If M (M> 2) line segments intersect at the same point, calculation should be repeated.
// You can assume two segments do not intersect more than one. (Do not overlap)

// very general solution: through each two line segments to determine whether there intersection.

// determine whether two lines intersect approach: two simultaneous linear. Seek intersection coordinates. . .

// high school knowledge, do not remember, can be derived directly.

// linear equation general formula: (y2 - y1) * x + (x1 - x2) * y + (x2 * y1 - x1 * y2) = 0; original formula (A * x + B * y + C = 0)

// intersection coordinates x: (b1 * c2 - b2 * c1) / (a1 * b2 - a2 * b1), y: (a2 * c1 - a1 * c2) / (a1 * b2 - a2 * b1);

// If a1 * b2 - a2 * b1 == 0 indicates that there is no intersection (parallel)

 1 #include<cstdio>
 2 
 3 struct Line {
 4     double x1, y1, x2, y2;
 5 } line[105];
 6 
 7 int n;
 8 
 9 bool intersect(int l1, int l2) {
10     double a1 = line[l1].y2 - line[l1].y1;
11     double b1 = line[l1].x1 - line[l1].x2;
12     double c1 = line[l1].x2*line[l1].y1 - line[l1].x1*line[l1].y2;
13     double a2 = line[l2].y2 - line[l2].y1;
14     double b2 = line[l2].x1 - line[l2].x2;
15     double c2 = line[l2].x2*line[l2].y1 - line[l2].x1*line[l2].y2;
16     if(a1*b2 - a2*b1 == 0) return false;
17     double x = (b1*c2 - b2*c1)/(a1*b2 - a2*b1);
18     double y = (a2*c1 - a1*c2)/(a1*b2 - a2*b1);
19     if((x >= line[l1].x1 && x <= line[l1].x2 || x <= line[l1].x1 && x >= line[l1].x2)
20     && (x >= line[l2].x1 && x <= line[l2].x2 || x <= line[l2].x1 && x >= line[l2].x2)) return true;
21     return false;
22 }
23 
24 int main() {
25     while(scanf("%d", &n) == 1 && n) {
26         for(int i = 0; i != n; ++i) {
27             scanf("%lf%lf%lf%lf", &line[i].x1, &line[i].y1, &line[i].x2, &line[i].y2);
28         }
29         int ans = 0;
30         for(int i = 0; i != n-1; ++i) {
31             for(int j = i+1; j != n; ++j) {
32                 if(intersect(i, j)) ans++;
33             }
34         }
35         printf("%d\n", ans);
36     }
37     return 0;
38 }