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hdu 1086

来源：华拓科技网

题目地址：

You can Solve a Geometry Problem too

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1746 Accepted Submission(s): 821
#include <iostream> using namespace std; #include <cmath> #include <cstdio> //跨立试验：如果两线段相交，则两线段必然相互跨立对方。 //若a1b1跨立a2b2，矢量(a1-a2)和(b1-a2)位于矢量(a2-b2) 的两侧 // i j i i i i i j //即 (a1-a2)×(a1-b1)*(a1-b1)×(a1-b2)>=0 //同理,a2b2跨立a1b1 (a2-a1)×(a2-b2)*(a2-b2)×(a2-b1)>=0 typedef struct { double x1,y1,x2,y2; }node; node t[105]; int n; int mul(double a,double b,double c,double d) { return a*d-b*c; } bool judge(int i,int j) { double a=mul(t[i].x1-t[j].x1,t[i].y1-t[j].y1,t[i].x1-t[i].x2,t[i].y1-t[i].y2); double b=mul(t[i].x1-t[i].x2,t[i].y1-t[i].y2,t[i].x1-t[j].x2,t[i].y1-t[j].y2); a=a*b; double c=mul(t[j].x1-t[i].x1,t[j].y1-t[i].y1,t[j].x1-t[j].x2,t[j].y1-t[j].y2); double d=mul(t[j].x1-t[j].x2,t[j].y1-t[j].y2,t[j].x1-t[i].x2,t[j].y1-t[i].y2); c=c*d; if(a>=0&&c>=0) return true; return false; } int main() { //freopen("1.in","r",stdin); while(scanf("%d",&n),n) { for(int i=1;i<=n;i++) scanf("%lf%lf%lf%lf",&t[i].x1,&t[i].y1,&t[i].x2,&t[i].y2); int count=0; for(int i=1;i<n;i++) { for(int j=i+1;j<=n;j++) if(judge(i,j)) count++; } cout<<count<<endl; } return 1; }

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
  
  
  
  
   
   这道题目搁在那边好久了，哎，考虑不周全
  
  
  
  
   
   于是今天花了些时间研究了下
  
  
  
  
   
   判断线段相交有个简单的方法，就是 跨立试验
  
  
  
  
   
   见：
   
   #判断点是否在线段上
  
  
  
  
   
   我的代码还可以优化，可以加上 快速排斥的方法

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