HDU1086判断线段相交

HDU1086

You can Solve a Geometry Problem too

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10924 Accepted Submission(s): 5393

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

　　可以把线段看成直线，计算出交点，看交点是不是在可行的范围内；注意一下斜率等于０的情况就行了；

 1 #include<algorithm>
 2 #include<iostream>
 3 #include<iomanip>
 4 #include<cstdlib>
 5 #include<cstring>
 6 #include<vector>
 7 #include<cstdio>
 8 #include<stack>
 9 #include<queue>
10 #include<cmath>
11 #include<ctime>
12 #include<set>
13 #include<map>
14 #define ll long long
15 #define rep(i,a,b) for(register int i=a;i<=b;i++)
16 #define inf 1<<29
17 #define re register
18 using namespace std;
19 #define eps 1e-6
20 const int N=110;
21 struct point{
22     double x,y;
23 }d[2];
24 struct line{
25     double k,b;
26     double l,r,ly,ry;
27 }L[N];
28 int n;
29 inline bool pd(double x,double y,int k) {
30     return (x>=L[k].l&&x<=L[k].r&&y>=L[k].ly&&y<=L[k].ry);
31 }
32 inline bool judge(int i,int j) {
33     if(fabs(L[i].k-L[j].k)<=eps) return 0;
34     if(fabs(L[i].k-0)<=eps&&fabs(L[j].k-0)<=eps) return 0;
35     if(fabs(L[i].k-0)>eps&&fabs(L[j].k-0)>eps) {
36         double x=(L[j].b-L[i].b)/(L[i].k-L[j].k),
37             y=L[i].k*x+L[i].b;
38         if(pd(x,y,i)&&pd(x,y,j)) return 1;
39         else return 0;
40     }
41     if(fabs(L[i].k-0)<=eps) {
42         double x=L[i].l;
43         double y=L[j].k*x+L[j].b;
44         return (y>=L[j].ly&&y<=L[j].ry);
45     }
46     double x=L[j].l;
47     double y=L[i].k*x+L[i].b;
48     return (y>=L[i].ly&&y<=L[i].ry);
49 }
50 int main() {
51     freopen("Y.in","r",stdin);
52     freopen("Y.out","w",stdout);
53     while(scanf("%d",&n)&&n) {
54         for(int i=1;i<=n;i++) {
55             scanf("%lf%lf%lf%lf",&d[0].x,&d[0].y,&d[1].x,&d[1].y);
56             L[i].l=min(d[0].x,d[1].x),L[i].r=max(d[0].x,d[1].x);
57             L[i].ly=min(d[0].y,d[1].y),L[i].ry=max(d[0].y,d[1].y);
58             if(fabs(d[1].x-d[0].x)<=eps)
59                 L[i].k=0;
60             else {
61                 L[i].k=(d[1].y-d[0].y)/(d[1].x-d[0].x);
62                 L[i].b=d[0].y-L[i].k*d[0].x;
63             }
64         }
65         int ans=0;
66         for(int i=1;i<n;i++)
67             for(int j=i+1;j<=n;j++)
68                 if(judge(i,j)) ans++;
69         cout<<ans<<endl;
70     }
71     return 0;
72 }