You can Solve a Geometry Problem too

                                        Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
                                        

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

直接O(N^2)判断两线段是否相交即可。

判断线段是否相交的模板：

 1 inline double CrossProduct(node a, node b, node c){
 2     return (b.x - a.x) * (c.y - a.y) - (b.y - a.y) * (c.x - a.x);
 3 }
 4 //Calculate the crossproduct
 5
 6
 7 inline bool SegX(node p1, node p2, node p3, node p4){
 8     double d1 = CrossProduct(p3, p4, p1);
 9     double d2 = CrossProduct(p3, p4, p2);
10     double d3 = CrossProduct(p1, p2, p3);
11     double d4 = CrossProduct(p1, p2, p4);
12     return (d1 * d2 <= 0 && d3 * d4 <= 0);
13 }
14 //Judge whether the line segments intersact

那么直接套用一下就好了。

 1 #include <bits/stdc++.h>
 2
 3 using namespace std;
 4
 5 struct node{
 6     double x, y;
 7 } pa[100010], pb[100010];
 8
 9 int n, num;
10
11 inline double CrossProduct(node a, node b, node c){
12     return (b.x - a.x) * (c.y - a.y) - (b.y - a.y) * (c.x - a.x);
13 }
14
15
16 inline bool SegX(node p1, node p2, node p3, node p4){
17     double d1 = CrossProduct(p3, p4, p1);
18     double d2 = CrossProduct(p3, p4, p2);
19     double d3 = CrossProduct(p1, p2, p3);
20     double d4 = CrossProduct(p1, p2, p4);
21     return (d1 * d2 <= 0 && d3 * d4 <= 0);
22 }
23
24 int main(){
25
26     while (~scanf("%d", &n), n){
27         num = 0;
28         for (int i = 1; i <= n; ++i) scanf("%lf%lf%lf%lf", &pa[i].x, &pa[i].y, &pb[i].x, &pb[i].y);
29         for (int i = 1; i <= n - 1; ++i)
30             for (int j = i + 1; j <= n; ++j)
31                 if (SegX(pa[i], pb[i], pa[j], pb[j])) ++num;
32         printf("%d\n", num);
33     }
34
35     return 0;
36
37 }