ZOJ 1648 线段相交

题意：

是否存在规范相交

 

View Code 
 1 #include <iostream>
 2 #include <cstdlib>
 3 #include <cstdio>
 4 #include <cstring>
 5 #include <algorithm>
 6 #include <cmath>
 7 
 8 #define N 2020
 9 #define EPS 1e-4
10 //规范相交 
11 using namespace std;
12 
13 struct PO
14 {
15     double x,y;
16 };
17 
18 struct LI
19 {
20     PO a,b;
21 }li[N];
22 
23 int n;
24 
25 inline int doublecmp(double x)
26 {
27     if(x>EPS) return 1;
28     else if(x<-EPS) return -1;
29     return 0;
30 }
31 
32 inline double cross(PO &a,PO &b,PO &c)
33 {
34     return (c.x-a.x)*(b.y-a.y)-(c.y-a.y)*(b.x-a.x);
35 }
36 
37 inline double dot(PO &a,PO &b,PO &c)
38 {
39     return (b.x-a.x)*(c.x-a.x)+(b.y-a.y)*(c.y-a.y);
40 }
41 
42 inline bool segcross(LI &a,LI &b)
43 {
44     int p1,p2,d1,d2;
45     //b跨立a 
46     p1=doublecmp(cross(a.a,a.b,b.a));
47     p2=doublecmp(cross(a.a,a.b,b.b));
48     //a跨立b 
49     d1=doublecmp(cross(b.a,b.b,a.a));
50     d2=doublecmp(cross(b.a,b.b,a.b));
51     if(p1*p2<0&&d1*d2<0) return true;
52     else return false;
53 }
54 
55 inline void read()
56 {
57     for(int i=1;i<=n;i++)
58         scanf("%lf%lf%lf%lf",&li[i].a.x,&li[i].a.y,&li[i].b.x,&li[i].b.y);
59 }
60 
61 inline void go()
62 {
63     for(int i=1;i<=n;i++)
64         for(int j=i+1;j<=n;j++)
65             if(segcross(li[i],li[j])) {puts("burned!");return;}
66     puts("ok!");
67 }
68 
69 int main()
70 {
71     while(scanf("%d",&n)!=EOF) read(),go();
72     return 0;
73 }

 

 

posted @ 2013-02-23 21:25  proverbs  阅读(206)  评论(0)  编辑  收藏  举报