HDU 1086 You can Solve a Geometry Problem too

You can Solve a Geometry Problem too

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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

20.00 0.00 1.00 1.000.00 1.00 1.00 0.0030.00 0.00 1.00 1.000.00 1.00 1.00 0.0000.00 0.00 1.00 0.000 

 

Sample Output

13 

 

  题意：给你n条线段，判断它们有多少个交点，交点需要重复计算。

  

         注意向量的点乘与叉乘

#include<stdio.h>struct xian{    double x1,y1;    double x2,y2;}f[101];bool find(xian a,xian b){if(((a.x1-b.x1)*(a.y2-b.y1)-(a.x2-b.x1)*(a.y1-b.y1))*((a.x1-b.x2)*(a.y2-b.y2)-(a.x2-b.x2)*(a.y1-b.y2))>0)return false;if(((b.x1-a.x1)*(b.y2-a.y1)-(b.x2-a.x1)*(b.y1-a.y1))*((b.x1-a.x2)*(b.y2-a.y2)-(b.x2-a.x2)*(b.y1-a.y2))>0)return false;return true;}int main(){    int n;    while(~scanf("%d",&n)&&n)    {        int i,j,t = 0;        for(i = 1 ; i <= n ; i ++)            scanf("%lf%lf%lf%lf",&f[i].x1,&f[i].y1,&f[i].x2,&f[i].y2);        for(i = 1 ; i < n ; i ++)            for(j = i+1 ; j <= n ; j ++)        {            if(find(f[i],f[j]))  //判定每一次是否相交                t++;        }        printf("%d\n",t);    }    return 0;}